diff --git a/leetcode/0009.Palindrome-Number/9. Palindrome Number.go b/leetcode/0009.Palindrome-Number/9. Palindrome Number.go new file mode 100644 index 00000000..dfc0935b --- /dev/null +++ b/leetcode/0009.Palindrome-Number/9. Palindrome Number.go @@ -0,0 +1,20 @@ +package leetcode + +import "strconv" + +func isPalindrome(x int) bool { + if x < 0 { + return false + } + if x < 10 { + return true + } + s := strconv.Itoa(x) + length := len(s) + for i := 0; i <= length/2; i++ { + if s[i] != s[length-1-i] { + return false + } + } + return true +} diff --git a/leetcode/0009.Palindrome-Number/9. Palindrome Number_test.go b/leetcode/0009.Palindrome-Number/9. Palindrome Number_test.go new file mode 100644 index 00000000..a9eaff03 --- /dev/null +++ b/leetcode/0009.Palindrome-Number/9. Palindrome Number_test.go @@ -0,0 +1,72 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question9 struct { + para9 + ans9 +} + +// para 是参数 +// one 代表第一个参数 +type para9 struct { + one int +} + +// ans 是答案 +// one 代表第一个答案 +type ans9 struct { + one bool +} + +func Test_Problem9(t *testing.T) { + + qs := []question9{ + + question9{ + para9{121}, + ans9{true}, + }, + + question9{ + para9{-121}, + ans9{false}, + }, + + question9{ + para9{10}, + ans9{false}, + }, + + question9{ + para9{321}, + ans9{false}, + }, + + question9{ + para9{-123}, + ans9{false}, + }, + + question9{ + para9{120}, + ans9{false}, + }, + + question9{ + para9{1534236469}, + ans9{false}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 9------------------------\n") + + for _, q := range qs { + _, p := q.ans9, q.para9 + fmt.Printf("【input】:%v 【output】:%v\n", p.one, isPalindrome(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0009.Palindrome-Number/README.md b/leetcode/0009.Palindrome-Number/README.md new file mode 100644 index 00000000..d01fc716 --- /dev/null +++ b/leetcode/0009.Palindrome-Number/README.md @@ -0,0 +1,69 @@ +# [9. Palindrome Number](https://leetcode.com/problems/palindrome-number/) + + +## 题目 + +Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. + +**Example 1**: + +``` +Input: 121 +Output: true +``` + +**Example 2**: + +``` +Input: -121 +Output: false +Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome. +``` + +**Example 3**: + +``` +Input: 10 +Output: false +Explanation: Reads 01 from right to left. Therefore it is not a palindrome. +``` + +**Follow up**: + +Coud you solve it without converting the integer to a string? + +## 题目大意 + +判断一个整数是否是回文数。回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。 + +## 解题思路 + +- 判断一个整数是不是回文数。 +- 简单题。注意会有负数的情况，负数，个位数，10 都不是回文数。其他的整数再按照回文的规则判断。 + +## 代码 + +```go + +package leetcode + +import "strconv" + +func isPalindrome(x int) bool { + if x < 0 { + return false + } + if x < 10 { + return true + } + s := strconv.Itoa(x) + length := len(s) + for i := 0; i <= length/2; i++ { + if s[i] != s[length-1-i] { + return false + } + } + return true +} + +``` \ No newline at end of file diff --git a/leetcode/0013.Roman-to-Integer/13. Roman to Integer.go b/leetcode/0013.Roman-to-Integer/13. Roman to Integer.go new file mode 100644 index 00000000..51fd4782 --- /dev/null +++ b/leetcode/0013.Roman-to-Integer/13. Roman to Integer.go @@ -0,0 +1,29 @@ +package leetcode + +var roman = map[string]int{ + "I": 1, + "V": 5, + "X": 10, + "L": 50, + "C": 100, + "D": 500, + "M": 1000, +} + +func romanToInt(s string) int { + if s == "" { + return 0 + } + num, lastint, total := 0, 0, 0 + for i := 0; i < len(s); i++ { + char := s[len(s)-(i+1) : len(s)-i] + num = roman[char] + if num < lastint { + total = total - num + } else { + total = total + num + } + lastint = num + } + return total +} diff --git a/leetcode/0013.Roman-to-Integer/13. Roman to Integer_test.go b/leetcode/0013.Roman-to-Integer/13. Roman to Integer_test.go new file mode 100644 index 00000000..539e6645 --- /dev/null +++ b/leetcode/0013.Roman-to-Integer/13. Roman to Integer_test.go @@ -0,0 +1,67 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question13 struct { + para13 + ans13 +} + +// para 是参数 +// one 代表第一个参数 +type para13 struct { + one string +} + +// ans 是答案 +// one 代表第一个答案 +type ans13 struct { + one int +} + +func Test_Problem13(t *testing.T) { + + qs := []question13{ + + question13{ + para13{"III"}, + ans13{3}, + }, + + question13{ + para13{"IV"}, + ans13{4}, + }, + + question13{ + para13{"IX"}, + ans13{9}, + }, + + question13{ + para13{"LVIII"}, + ans13{58}, + }, + + question13{ + para13{"MCMXCIV"}, + ans13{1994}, + }, + + question13{ + para13{"MCMXICIVI"}, + ans13{2014}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 13------------------------\n") + + for _, q := range qs { + _, p := q.ans13, q.para13 + fmt.Printf("【input】:%v 【output】:%v\n", p.one, romanToInt(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0013.Roman-to-Integer/README.md b/leetcode/0013.Roman-to-Integer/README.md new file mode 100644 index 00000000..44d758eb --- /dev/null +++ b/leetcode/0013.Roman-to-Integer/README.md @@ -0,0 +1,132 @@ +# [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer/) + + +## 题目 + +Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`. + +``` +Symbol Value +I 1 +V 5 +X 10 +L 50 +C 100 +D 500 +M 1000 +``` + +For example, two is written as `II` in Roman numeral, just two one's added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`. + +Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used: + +- `I` can be placed before `V` (5) and `X` (10) to make 4 and 9. +- `X` can be placed before `L` (50) and `C` (100) to make 40 and 90. +- `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900. + +Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. + +**Example 1**: + +``` +Input: "III" +Output: 3 +``` + +**Example 2**: + +``` +Input: "IV" +Output: 4 +``` + +**Example 3**: + +``` +Input: "IX" +Output: 9 +``` + +**Example 4**: + +``` +Input: "LVIII" +Output: 58 +Explanation: L = 50, V= 5, III = 3. +``` + +**Example 5**: + +``` +Input: "MCMXCIV" +Output: 1994 +Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. +``` + +## 题目大意 + +罗马数字包含以下七种字符: I， V， X， L，C，D 和 M。 + +```go + +字符 数值 +I 1 +V 5 +X 10 +L 50 +C 100 +D 500 +M 1000 + +``` + +例如， 罗马数字 2 写做 II ，即为两个并列的 1。12 写做 XII ，即为 X + II 。 27 写做  XXVII, 即为 XX + V + II 。 + +通常情况下，罗马数字中小的数字在大的数字的右边。但也存在特例，例如 4 不写做 IIII，而是 IV。数字 1 在数字 5 的左边，所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地，数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况： + +- I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。 +- X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。  +- C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。 + +给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。 + +## 解题思路 + +- 给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。 +- 简单题。按照题目中罗马数字的字符数值，计算出对应罗马数字的十进制数即可。 + +## 代码 + +```go + +package leetcode + +var roman = map[string]int{ + "I": 1, + "V": 5, + "X": 10, + "L": 50, + "C": 100, + "D": 500, + "M": 1000, +} + +func romanToInt(s string) int { + if s == "" { + return 0 + } + num, lastint, total := 0, 0, 0 + for i := 0; i < len(s); i++ { + char := s[len(s)-(i+1) : len(s)-i] + num = roman[char] + if num < lastint { + total = total - num + } else { + total = total + num + } + lastint = num + } + return total +} + +``` \ No newline at end of file diff --git a/leetcode/0067.Add-Binary/67. Add Binary.go b/leetcode/0067.Add-Binary/67. Add Binary.go new file mode 100644 index 00000000..8d22cf04 --- /dev/null +++ b/leetcode/0067.Add-Binary/67. Add Binary.go @@ -0,0 +1,38 @@ +package leetcode + +import ( + "strconv" + "strings" +) + +func addBinary(a string, b string) string { + if len(b) > len(a) { + a, b = b, a + } + + res := make([]string, len(a)+1) + i, j, k, c := len(a)-1, len(b)-1, len(a), 0 + for i >= 0 && j >= 0 { + ai, _ := strconv.Atoi(string(a[i])) + bj, _ := strconv.Atoi(string(b[j])) + res[k] = strconv.Itoa((ai + bj + c) % 2) + c = (ai + bj + c) / 2 + i-- + j-- + k-- + } + + for i >= 0 { + ai, _ := strconv.Atoi(string(a[i])) + res[k] = strconv.Itoa((ai + c) % 2) + c = (ai + c) / 2 + i-- + k-- + } + + if c > 0 { + res[k] = strconv.Itoa(c) + } + + return strings.Join(res, "") +} diff --git a/leetcode/0067.Add-Binary/67. Add Binary_test.go b/leetcode/0067.Add-Binary/67. Add Binary_test.go new file mode 100644 index 00000000..50746ef5 --- /dev/null +++ b/leetcode/0067.Add-Binary/67. Add Binary_test.go @@ -0,0 +1,48 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question67 struct { + para67 + ans67 +} + +// para 是参数 +// one 代表第一个参数 +type para67 struct { + a string + b string +} + +// ans 是答案 +// one 代表第一个答案 +type ans67 struct { + one string +} + +func Test_Problem67(t *testing.T) { + + qs := []question67{ + + question67{ + para67{"11", "1"}, + ans67{"100"}, + }, + + question67{ + para67{"1010", "1011"}, + ans67{"10101"}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 67------------------------\n") + + for _, q := range qs { + _, p := q.ans67, q.para67 + fmt.Printf("【input】:%v 【output】:%v\n", p, addBinary(p.a, p.b)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0067.Add-Binary/README.md b/leetcode/0067.Add-Binary/README.md new file mode 100644 index 00000000..91d977e5 --- /dev/null +++ b/leetcode/0067.Add-Binary/README.md @@ -0,0 +1,76 @@ +# [67. Add Binary](https://leetcode.com/problems/add-binary/) + + +## 题目 + +Given two binary strings, return their sum (also a binary string). + +The input strings are both **non-empty** and contains only characters `1` or `0`. + +**Example 1**: + +``` +Input: a = "11", b = "1" +Output: "100" +``` + +**Example 2**: + +``` +Input: a = "1010", b = "1011" +Output: "10101" +``` + +## 题目大意 + +给你两个二进制字符串，返回它们的和（用二进制表示）。输入为 非空 字符串且只包含数字 1 和 0。 + +## 解题思路 + +- 要求输出 2 个二进制数的和，结果也用二进制表示。 +- 简单题。按照二进制的加法规则做加法即可。 + +## 代码 + +```go + +package leetcode + +import ( + "strconv" + "strings" +) + +func addBinary(a string, b string) string { + if len(b) > len(a) { + a, b = b, a + } + + res := make([]string, len(a)+1) + i, j, k, c := len(a)-1, len(b)-1, len(a), 0 + for i >= 0 && j >= 0 { + ai, _ := strconv.Atoi(string(a[i])) + bj, _ := strconv.Atoi(string(b[j])) + res[k] = strconv.Itoa((ai + bj + c) % 2) + c = (ai + bj + c) / 2 + i-- + j-- + k-- + } + + for i >= 0 { + ai, _ := strconv.Atoi(string(a[i])) + res[k] = strconv.Itoa((ai + c) % 2) + c = (ai + c) / 2 + i-- + k-- + } + + if c > 0 { + res[k] = strconv.Itoa(c) + } + + return strings.Join(res, "") +} + +``` \ No newline at end of file diff --git a/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title.go b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title.go new file mode 100644 index 00000000..e9db5f7c --- /dev/null +++ b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title.go @@ -0,0 +1,13 @@ +package leetcode + +func convertToTitle(n int) string { + result := []byte{} + for n > 0 { + result = append(result, 'A'+byte((n-1)%26)) + n = (n - 1) / 26 + } + for i, j := 0, len(result)-1; i < j; i, j = i+1, j-1 { + result[i], result[j] = result[j], result[i] + } + return string(result) +} diff --git a/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title_test.go b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title_test.go new file mode 100644 index 00000000..7b9ef119 --- /dev/null +++ b/leetcode/0168.Excel-Sheet-Column-Title/168. Excel Sheet Column Title_test.go @@ -0,0 +1,67 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question168 struct { + para168 + ans168 +} + +// para 是参数 +// one 代表第一个参数 +type para168 struct { + n int +} + +// ans 是答案 +// one 代表第一个答案 +type ans168 struct { + one string +} + +func Test_Problem168(t *testing.T) { + + qs := []question168{ + + question168{ + para168{1}, + ans168{"A"}, + }, + + question168{ + para168{28}, + ans168{"AB"}, + }, + + question168{ + para168{701}, + ans168{"ZY"}, + }, + + question168{ + para168{10011}, + ans168{"NUA"}, + }, + + question168{ + para168{999}, + ans168{"ALK"}, + }, + + question168{ + para168{681}, + ans168{"ZE"}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 168------------------------\n") + + for _, q := range qs { + _, p := q.ans168, q.para168 + fmt.Printf("【input】:%v 【output】:%v\n", p, convertToTitle(p.n)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0168.Excel-Sheet-Column-Title/README.md b/leetcode/0168.Excel-Sheet-Column-Title/README.md new file mode 100644 index 00000000..c1ecdd58 --- /dev/null +++ b/leetcode/0168.Excel-Sheet-Column-Title/README.md @@ -0,0 +1,80 @@ +# [168. Excel Sheet Column Title](https://leetcode.com/problems/excel-sheet-column-title/) + +## 题目 + +Given a positive integer, return its corresponding column title as appear in an Excel sheet. + +For example: + +``` + 1 -> A + 2 -> B + 3 -> C + ... + 26 -> Z + 27 -> AA + 28 -> AB + ... +``` + +**Example 1**: + +``` +Input: 1 +Output: "A" +``` + +**Example 2**: + +``` +Input: 28 +Output: "AB" +``` + +**Example 3**: + +``` +Input: 701 +Output: "ZY" +``` + +## 题目大意 + +给定一个正整数，返回它在 Excel 表中相对应的列名称。 + +例如， + + 1 -> A + 2 -> B + 3 -> C + ... + 26 -> Z + 27 -> AA + 28 -> AB + ... + + +## 解题思路 + +- 给定一个正整数，返回它在 Excel 表中的对应的列名称 +- 简单题。这一题就类似短除法的计算过程。以 26 进制的字母编码。按照短除法先除，然后余数逆序输出即可。 + +## 代码 + +```go + +package leetcode + +func convertToTitle(n int) string { + result := []byte{} + for n > 0 { + result = append(result, 'A'+byte((n-1)%26)) + n = (n - 1) / 26 + } + for i, j := 0, len(result)-1; i < j; i, j = i+1, j-1 { + result[i], result[j] = result[j], result[i] + } + return string(result) +} + +``` \ No newline at end of file diff --git a/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number.go b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number.go new file mode 100644 index 00000000..5bcfed11 --- /dev/null +++ b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number.go @@ -0,0 +1,10 @@ +package leetcode + +func titleToNumber(s string) int { + val, res := 0, 0 + for i := 0; i < len(s); i++ { + val = int(s[i] - 'A' + 1) + res = res*26 + val + } + return res +} diff --git a/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number_test.go b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number_test.go new file mode 100644 index 00000000..955b4701 --- /dev/null +++ b/leetcode/0171.Excel-Sheet-Column-Number/171. Excel Sheet Column Number_test.go @@ -0,0 +1,57 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question171 struct { + para171 + ans171 +} + +// para 是参数 +// one 代表第一个参数 +type para171 struct { + s string +} + +// ans 是答案 +// one 代表第一个答案 +type ans171 struct { + one int +} + +func Test_Problem171(t *testing.T) { + + qs := []question171{ + + question171{ + para171{"A"}, + ans171{1}, + }, + + question171{ + para171{"AB"}, + ans171{28}, + }, + + question171{ + para171{"ZY"}, + ans171{701}, + }, + + question171{ + para171{"ABC"}, + ans171{731}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 171------------------------\n") + + for _, q := range qs { + _, p := q.ans171, q.para171 + fmt.Printf("【input】:%v 【output】:%v\n", p, titleToNumber(p.s)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0171.Excel-Sheet-Column-Number/README.md b/leetcode/0171.Excel-Sheet-Column-Number/README.md new file mode 100644 index 00000000..8c9f6651 --- /dev/null +++ b/leetcode/0171.Excel-Sheet-Column-Number/README.md @@ -0,0 +1,67 @@ +# [171. Excel Sheet Column Number](https://leetcode.com/problems/excel-sheet-column-number/) + + +## 题目 + +Given a column title as appear in an Excel sheet, return its corresponding column number. + +For example: + +``` + A -> 1 + B -> 2 + C -> 3 + ... + Z -> 26 + AA -> 27 + AB -> 28 + ... +``` + +**Example 1**: + +``` +Input: "A" +Output: 1 +``` + +**Example 2**: + +``` +Input: "AB" +Output: 28 +``` + +**Example 3**: + +``` +Input: "ZY" +Output: 701 +``` + +## 题目大意 + +给定一个 Excel 表格中的列名称，返回其相应的列序号。 + + +## 解题思路 + +- 给出 Excel 中列的名称，输出其对应的列序号。 +- 简单题。这一题是第 168 题的逆序题。按照 26 进制还原成十进制即可。 + +## 代码 + +```go + +package leetcode + +func titleToNumber(s string) int { + val, res := 0, 0 + for i := 0; i < len(s); i++ { + val = int(s[i] - 'A' + 1) + res = res*26 + val + } + return res +} + +``` \ No newline at end of file diff --git a/leetcode/0258.Add-Digits/258. Add Digits.go b/leetcode/0258.Add-Digits/258. Add Digits.go new file mode 100644 index 00000000..b64893b8 --- /dev/null +++ b/leetcode/0258.Add-Digits/258. Add Digits.go @@ -0,0 +1,13 @@ +package leetcode + +func addDigits(num int) int { + for num > 9 { + cur := 0 + for num != 0 { + cur += num % 10 + num /= 10 + } + num = cur + } + return num +} diff --git a/leetcode/0258.Add-Digits/258. Add Digits_test.go b/leetcode/0258.Add-Digits/258. Add Digits_test.go new file mode 100644 index 00000000..78754617 --- /dev/null +++ b/leetcode/0258.Add-Digits/258. Add Digits_test.go @@ -0,0 +1,52 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question258 struct { + para258 + ans258 +} + +// para 是参数 +// one 代表第一个参数 +type para258 struct { + one int +} + +// ans 是答案 +// one 代表第一个答案 +type ans258 struct { + one int +} + +func Test_Problem258(t *testing.T) { + + qs := []question258{ + + question258{ + para258{38}, + ans258{2}, + }, + + question258{ + para258{88}, + ans258{7}, + }, + + question258{ + para258{96}, + ans258{6}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 258------------------------\n") + + for _, q := range qs { + _, p := q.ans258, q.para258 + fmt.Printf("【input】:%v 【output】:%v\n", p, addDigits(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0258.Add-Digits/README.md b/leetcode/0258.Add-Digits/README.md new file mode 100644 index 00000000..8c811904 --- /dev/null +++ b/leetcode/0258.Add-Digits/README.md @@ -0,0 +1,47 @@ +# [258. Add Digits](https://leetcode.com/problems/add-digits/) + + +## 题目 + +Given a non-negative integer `num`, repeatedly add all its digits until the result has only one digit. + +**Example**: + +``` +Input: 38 +Output: 2 +Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. + Since 2 has only one digit, return it. +``` + +**Follow up**: Could you do it without any loop/recursion in O(1) runtime? + +## 题目大意 + +给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。 + + +## 解题思路 + +- 给定一个非负整数，反复加各个位上的数，直到结果为一位数为止，最后输出这一位数。 +- 简单题。按照题意循环累加即可。 + +## 代码 + +```go + +package leetcode + +func addDigits(num int) int { + for num > 9 { + cur := 0 + for num != 0 { + cur += num % 10 + num /= 10 + } + num = cur + } + return num +} + +``` \ No newline at end of file diff --git a/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements.go b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements.go new file mode 100644 index 00000000..fbdfeb29 --- /dev/null +++ b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements.go @@ -0,0 +1,14 @@ +package leetcode + +import "math" + +func minMoves(nums []int) int { + sum, min, l := 0, math.MaxInt32, len(nums) + for _, v := range nums { + sum += v + if min > v { + min = v + } + } + return sum - min*l +} diff --git a/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements_test.go b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements_test.go new file mode 100644 index 00000000..6a577b0f --- /dev/null +++ b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/453. Minimum Moves to Equal Array Elements_test.go @@ -0,0 +1,48 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question453 struct { + para453 + ans453 +} + +// para 是参数 +// one 代表第一个参数 +type para453 struct { + one []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans453 struct { + one int +} + +func Test_Problem453(t *testing.T) { + + qs := []question453{ + + question453{ + para453{[]int{4, 3, 2, 7, 8, 2, 3, 1}}, + ans453{22}, + }, + + question453{ + para453{[]int{1, 2, 3}}, + ans453{3}, + }, + // 如需多个测试，可以复制上方元素。 + } + + fmt.Printf("------------------------Leetcode Problem 453------------------------\n") + + for _, q := range qs { + _, p := q.ans453, q.para453 + fmt.Printf("【input】:%v 【output】:%v\n", p, minMoves(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/README.md b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/README.md new file mode 100644 index 00000000..280f22a2 --- /dev/null +++ b/leetcode/0453.Minimum-Moves-to-Equal-Array-Elements/README.md @@ -0,0 +1,51 @@ +# [453. Minimum Moves to Equal Array Elements](https://leetcode.com/problems/minimum-moves-to-equal-array-elements/) + + +## 题目 + +Given a **non-empty** integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. + +**Example**: + +``` +Input: +[1,2,3] + +Output: +3 + +Explanation: +Only three moves are needed (remember each move increments two elements): + +[1,2,3] => [2,3,3] => [3,4,3] => [4,4,4] +``` + +## 题目大意 + +给定一个长度为 n 的非空整数数组，找到让数组所有元素相等的最小移动次数。每次移动将会使 n - 1 个元素增加 1。 + +## 解题思路 + +- 给定一个数组，要求输出让所有元素都相等的最小步数。每移动一步都会使得 n - 1 个元素 + 1 。 +- 数学题。这道题正着思考会考虑到排序或者暴力的方法上去。反过来思考一下，使得每个元素都相同，意思让所有元素的差异变为 0 。每次移动的过程中，都有 n - 1 个元素 + 1，那么没有 + 1 的那个元素和其他 n - 1 个元素相对差异就缩小了。所以这道题让所有元素都变为相等的最少步数，即等于让所有元素相对差异减少到最小的那个数。想到这里，此题就可以优雅的解出来了。 + +## 代码 + +```go + +package leetcode + +import "math" + +func minMoves(nums []int) int { + sum, min, l := 0, math.MaxInt32, len(nums) + for _, v := range nums { + sum += v + if min > v { + min = v + } + } + return sum - min*l +} + +``` \ No newline at end of file diff --git a/leetcode/0507.Perfect-Number/507. Perfect Number.go b/leetcode/0507.Perfect-Number/507. Perfect Number.go new file mode 100644 index 00000000..ffeb6fac --- /dev/null +++ b/leetcode/0507.Perfect-Number/507. Perfect Number.go @@ -0,0 +1,24 @@ +package leetcode + +import "math" + +// 方法一 +func checkPerfectNumber(num int) bool { + if num <= 1 { + return false + } + sum, bound := 1, int(math.Sqrt(float64(num)))+1 + for i := 2; i < bound; i++ { + if num%i != 0 { + continue + } + corrDiv := num / i + sum += corrDiv + i + } + return sum == num +} + +// 方法二 打表 +func checkPerfectNumber_(num int) bool { + return num == 6 || num == 28 || num == 496 || num == 8128 || num == 33550336 +} diff --git a/leetcode/0507.Perfect-Number/507. Perfect Number_test.go b/leetcode/0507.Perfect-Number/507. Perfect Number_test.go new file mode 100644 index 00000000..20aebcbd --- /dev/null +++ b/leetcode/0507.Perfect-Number/507. Perfect Number_test.go @@ -0,0 +1,53 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question507 struct { + para507 + ans507 +} + +// para 是参数 +// one 代表第一个参数 +type para507 struct { + num int +} + +// ans 是答案 +// one 代表第一个答案 +type ans507 struct { + one bool +} + +func Test_Problem507(t *testing.T) { + + qs := []question507{ + + question507{ + para507{28}, + ans507{true}, + }, + + question507{ + para507{496}, + ans507{true}, + }, + + question507{ + para507{500}, + ans507{false}, + }, + // 如需多个测试，可以复制上方元素。 + } + + fmt.Printf("------------------------Leetcode Problem 507------------------------\n") + + for _, q := range qs { + _, p := q.ans507, q.para507 + fmt.Printf("【input】:%v 【output】:%v\n", p, checkPerfectNumber(p.num)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0507.Perfect-Number/README.md b/leetcode/0507.Perfect-Number/README.md new file mode 100644 index 00000000..12d0c764 --- /dev/null +++ b/leetcode/0507.Perfect-Number/README.md @@ -0,0 +1,64 @@ +# [507. Perfect Number](https://leetcode.com/problems/perfect-number/) + + + +## 题目 + +We define the Perfect Number is a **positive** integer that is equal to the sum of all its **positive** divisors except itself. + +Now, given an + +**integer** + +n, write a function that returns true when it is a perfect number and false when it is not. + +**Example**: + +``` +Input: 28 +Output: True +Explanation: 28 = 1 + 2 + 4 + 7 + 14 +``` + +**Note**: The input number **n** will not exceed 100,000,000. (1e8) + +## 题目大意 + +对于一个 正整数，如果它和除了它自身以外的所有正因子之和相等，我们称它为“完美数”。给定一个 整数 n， 如果他是完美数，返回 True，否则返回 False + +## 解题思路 + +- 给定一个整数，要求判断这个数是不是完美数。整数的取值范围小于 1e8 。 +- 简单题。按照题意描述，先获取这个整数的所有正因子，如果正因子的和等于原来这个数，那么它就是完美数。 +- 这一题也可以打表，1e8 以下的完美数其实并不多，就 5 个。 + +## 代码 + +```go + +package leetcode + +import "math" + +// 方法一 +func checkPerfectNumber(num int) bool { + if num <= 1 { + return false + } + sum, bound := 1, int(math.Sqrt(float64(num)))+1 + for i := 2; i < bound; i++ { + if num%i != 0 { + continue + } + corrDiv := num / i + sum += corrDiv + i + } + return sum == num +} + +// 方法二 打表 +func checkPerfectNumber_(num int) bool { + return num == 6 || num == 28 || num == 496 || num == 8128 || num == 33550336 +} + +``` \ No newline at end of file diff --git a/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication.go b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication.go new file mode 100644 index 00000000..d552b3e1 --- /dev/null +++ b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication.go @@ -0,0 +1,21 @@ +package leetcode + +import ( + "strconv" + "strings" +) + +func complexNumberMultiply(a string, b string) string { + realA, imagA := parse(a) + realB, imagB := parse(b) + real := realA*realB - imagA*imagB + imag := realA*imagB + realB*imagA + return strconv.Itoa(real) + "+" + strconv.Itoa(imag) + "i" +} + +func parse(s string) (int, int) { + ss := strings.Split(s, "+") + r, _ := strconv.Atoi(ss[0]) + i, _ := strconv.Atoi(ss[1][:len(ss[1])-1]) + return r, i +} diff --git a/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication_test.go b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication_test.go new file mode 100644 index 00000000..35ee1d1f --- /dev/null +++ b/leetcode/0537.Complex-Number-Multiplication/537. Complex Number Multiplication_test.go @@ -0,0 +1,48 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question537 struct { + para537 + ans537 +} + +// para 是参数 +// one 代表第一个参数 +type para537 struct { + a string + b string +} + +// ans 是答案 +// one 代表第一个答案 +type ans537 struct { + one string +} + +func Test_Problem537(t *testing.T) { + + qs := []question537{ + + question537{ + para537{"1+1i", "1+1i"}, + ans537{"0+2i"}, + }, + + question537{ + para537{"1+-1i", "1+-1i"}, + ans537{"0+-2i"}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 537------------------------\n") + + for _, q := range qs { + _, p := q.ans537, q.para537 + fmt.Printf("【input】:%v 【output】:%v\n", p, complexNumberMultiply(p.a, p.b)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0537.Complex-Number-Multiplication/README.md b/leetcode/0537.Complex-Number-Multiplication/README.md new file mode 100644 index 00000000..568404ba --- /dev/null +++ b/leetcode/0537.Complex-Number-Multiplication/README.md @@ -0,0 +1,73 @@ +# [537. Complex Number Multiplication](https://leetcode.com/problems/complex-number-multiplication/) + + +## 题目 + +Given two strings representing two [complex numbers](https://en.wikipedia.org/wiki/Complex_number). + +You need to return a string representing their multiplication. Note i2 = -1 according to the definition. + +**Example 1**: + +``` +Input: "1+1i", "1+1i" +Output: "0+2i" +Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i. +``` + +**Example 2**: + +``` +Input: "1+-1i", "1+-1i" +Output: "0+-2i" +Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i. +``` + +**Note**: + +1. The input strings will not have extra blank. +2. The input strings will be given in the form of **a+bi**, where the integer **a** and **b** will both belong to the range of [-100, 100]. And **the output should be also in this form**. + +## 题目大意 + +给定两个表示复数的字符串。返回表示它们乘积的字符串。注意，根据定义 i^2 = -1 。 + +注意: + +- 输入字符串不包含额外的空格。 +- 输入字符串将以 a+bi 的形式给出，其中整数 a 和 b 的范围均在 [-100, 100] 之间。输出也应当符合这种形式。 + + + +## 解题思路 + +- 给定 2 个字符串，要求这两个复数的乘积，输出也是字符串格式。 +- 数学题。按照复数的运算法则，i^2 = -1，最后输出字符串结果即可。 + +## 代码 + +```go + +package leetcode + +import ( + "strconv" + "strings" +) + +func complexNumberMultiply(a string, b string) string { + realA, imagA := parse(a) + realB, imagB := parse(b) + real := realA*realB - imagA*imagB + imag := realA*imagB + realB*imagA + return strconv.Itoa(real) + "+" + strconv.Itoa(imag) + "i" +} + +func parse(s string) (int, int) { + ss := strings.Split(s, "+") + r, _ := strconv.Atoi(ss[0]) + i, _ := strconv.Atoi(ss[1][:len(ss[1])-1]) + return r, i +} + +``` \ No newline at end of file diff --git a/leetcode/0561.Array-Partition-I/561. Array Partition I.go b/leetcode/0561.Array-Partition-I/561. Array Partition I.go new file mode 100644 index 00000000..c566c01b --- /dev/null +++ b/leetcode/0561.Array-Partition-I/561. Array Partition I.go @@ -0,0 +1,19 @@ +package leetcode + +func arrayPairSum(nums []int) int { + array := [20001]int{} + for i := 0; i < len(nums); i++ { + array[nums[i]+10000]++ + } + flag, sum := true, 0 + for i := 0; i < len(array); i++ { + for array[i] > 0 { + if flag { + sum = sum + i - 10000 + } + flag = !flag + array[i]-- + } + } + return sum +} diff --git a/leetcode/0561.Array-Partition-I/561. Array Partition I_test.go b/leetcode/0561.Array-Partition-I/561. Array Partition I_test.go new file mode 100644 index 00000000..ca062266 --- /dev/null +++ b/leetcode/0561.Array-Partition-I/561. Array Partition I_test.go @@ -0,0 +1,49 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question561 struct { + para561 + ans561 +} + +// para 是参数 +// one 代表第一个参数 +type para561 struct { + nums []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans561 struct { + one int +} + +func Test_Problem561(t *testing.T) { + + qs := []question561{ + + question561{ + para561{[]int{}}, + ans561{0}, + }, + + question561{ + para561{[]int{1, 4, 3, 2}}, + ans561{4}, + }, + + // 如需多个测试，可以复制上方元素。 + } + + fmt.Printf("------------------------Leetcode Problem 561------------------------\n") + + for _, q := range qs { + _, p := q.ans561, q.para561 + fmt.Printf("【input】:%v 【output】:%v\n", p, arrayPairSum(p.nums)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0561.Array-Partition-I/README.md b/leetcode/0561.Array-Partition-I/README.md new file mode 100644 index 00000000..d72ea633 --- /dev/null +++ b/leetcode/0561.Array-Partition-I/README.md @@ -0,0 +1,58 @@ +# [561. Array Partition I](https://leetcode.com/problems/array-partition-i/) + + +## 题目 + +Given an array of **2n** integers, your task is to group these integers into **n** pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible. + +**Example 1**: + +``` +Input: [1,4,3,2] + +Output: 4 +Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4). +``` + +**Note**: + +1. **n** is a positive integer, which is in the range of [1, 10000]. +2. All the integers in the array will be in the range of [-10000, 10000]. + +## 题目大意 + +给定长度为 2n 的数组, 你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得从1 到 n 的 min(ai, bi) 总和最大。 + + +## 解题思路 + +- 给定一个 2n 个数组，要求把它们分为 n 组一行，求出各组最小值的总和的最大值。 +- 由于题目给的数据范围不大，[-10000, 10000]，所以我们可以考虑用一个哈希表数组，里面存储 i - 10000 元素的频次，偏移量是 10000。这个哈希表能按递增的顺序访问数组，这样可以减少排序的耗时。题目要求求出分组以后求和的最大值，那么所有偏小的元素尽量都安排在一组里面，这样取 min 以后，对最大和影响不大。例如，(1 , 1) 这样安排在一起，min 以后就是 1 。但是如果把相差很大的两个元素安排到一起，那么较大的那个元素就“牺牲”了。例如，(1 , 10000)，取 min 以后就是 1，于是 10000 就“牺牲”了。所以需要优先考虑较小值。 +- 较小值出现的频次可能是奇数也可能是偶数。如果是偶数，那比较简单，把它们俩俩安排在一起就可以了。如果是奇数，那么它会落单一次，落单的那个需要和距离它最近的一个元素进行配对，这样对最终的和影响最小。较小值如果是奇数，那么就会影响后面元素的选择，后面元素如果是偶数，由于需要一个元素和前面的较小值配对，所以它剩下的又是奇数个。这个影响会依次传递到后面。所以用一个 flag 标记，如果当前集合中有剩余元素将被再次考虑，则此标志设置为 1。在从下一组中选择元素时，会考虑已考虑的相同额外元素。 +- 最后扫描过程中动态的维护 sum 值就可以了。 + +## 代码 + +```go + +package leetcode + +func arrayPairSum(nums []int) int { + array := [20001]int{} + for i := 0; i < len(nums); i++ { + array[nums[i]+10000]++ + } + flag, sum := true, 0 + for i := 0; i < len(array); i++ { + for array[i] > 0 { + if flag { + sum = sum + i - 10000 + } + flag = !flag + array[i]-- + } + } + return sum +} + +``` \ No newline at end of file diff --git a/leetcode/0598.Range-Addition-II/598. Range Addition II.go b/leetcode/0598.Range-Addition-II/598. Range Addition II.go new file mode 100644 index 00000000..9b58fd80 --- /dev/null +++ b/leetcode/0598.Range-Addition-II/598. Range Addition II.go @@ -0,0 +1,17 @@ +package leetcode + +func maxCount(m int, n int, ops [][]int) int { + minM, minN := m, n + for _, op := range ops { + minM = min(minM, op[0]) + minN = min(minN, op[1]) + } + return minM * minN +} + +func min(a, b int) int { + if a < b { + return a + } + return b +} diff --git a/leetcode/0598.Range-Addition-II/598. Range Addition II_test.go b/leetcode/0598.Range-Addition-II/598. Range Addition II_test.go new file mode 100644 index 00000000..0800f5f9 --- /dev/null +++ b/leetcode/0598.Range-Addition-II/598. Range Addition II_test.go @@ -0,0 +1,44 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question598 struct { + para598 + ans598 +} + +// para 是参数 +// one 代表第一个参数 +type para598 struct { + m int + n int + ops [][]int +} + +// ans 是答案 +// one 代表第一个答案 +type ans598 struct { + one int +} + +func Test_Problem598(t *testing.T) { + + qs := []question598{ + + question598{ + para598{3, 3, [][]int{[]int{2, 2}, []int{3, 3}}}, + ans598{4}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 598------------------------\n") + + for _, q := range qs { + _, p := q.ans598, q.para598 + fmt.Printf("【input】:%v 【output】:%v\n", p, maxCount(p.m, p.n, p.ops)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0598.Range-Addition-II/README.md b/leetcode/0598.Range-Addition-II/README.md new file mode 100644 index 00000000..a4606f95 --- /dev/null +++ b/leetcode/0598.Range-Addition-II/README.md @@ -0,0 +1,82 @@ +# [598. Range Addition II](https://leetcode.com/problems/range-addition-ii/) + + +## 题目 + +Given an m * n matrix **M** initialized with all **0**'s and several update operations. + +Operations are represented by a 2D array, and each operation is represented by an array with two **positive** integers **a** and **b**, which means **M[i][j]** should be **added by one** for all **0 <= i < a** and **0 <= j < b**. + +You need to count and return the number of maximum integers in the matrix after performing all the operations. + +**Example 1**: + +``` +Input: +m = 3, n = 3 +operations = [[2,2],[3,3]] +Output: 4 +Explanation: +Initially, M = +[[0, 0, 0], + [0, 0, 0], + [0, 0, 0]] + +After performing [2,2], M = +[[1, 1, 0], + [1, 1, 0], + [0, 0, 0]] + +After performing [3,3], M = +[[2, 2, 1], + [2, 2, 1], + [1, 1, 1]] + +So the maximum integer in M is 2, and there are four of it in M. So return 4. +``` + +**Note**: + +1. The range of m and n is [1,40000]. +2. The range of a is [1,m], and the range of b is [1,n]. +3. The range of operations size won't exceed 10,000. + +## 题目大意 + +给定一个初始元素全部为 0，大小为 m*n 的矩阵 M 以及在 M 上的一系列更新操作。操作用二维数组表示，其中的每个操作用一个含有两个正整数 a 和 b 的数组表示，含义是将所有符合 0 <= i < a 以及 0 <= j < b 的元素 M[i][j] 的值都增加 1。在执行给定的一系列操作后，你需要返回矩阵中含有最大整数的元素个数。 + +注意: + +- m 和 n 的范围是 [1,40000]。 +- a 的范围是 [1,m]，b 的范围是 [1,n]。 +- 操作数目不超过 10000。 + + +## 解题思路 + +- 给定一个初始都为 0 的 m * n 的矩阵，和一个操作数组。经过一系列的操作以后，最终输出矩阵中最大整数的元素个数。每次操作都使得一个矩形内的元素都 + 1 。 +- 这一题乍一看像线段树的区间覆盖问题，但是实际上很简单。如果此题是任意的矩阵，那就可能用到线段树了。这一题每个矩阵的起点都包含 [0 , 0] 这个元素，也就是说每次操作都会影响第一个元素。那么这道题就很简单了。经过 n 次操作以后，被覆盖次数最多的矩形区间，一定就是最大整数所在的区间。由于起点都是第一个元素，所以我们只用关心矩形的右下角那个坐标。右下角怎么计算呢？只用每次动态的维护一下矩阵长和宽的最小值即可。 + +## 代码 + +```go + +package leetcode + +func maxCount(m int, n int, ops [][]int) int { + minM, minN := m, n + for _, op := range ops { + minM = min(minM, op[0]) + minN = min(minN, op[1]) + } + return minM * minN +} + +func min(a, b int) int { + if a < b { + return a + } + return b +} + +``` \ No newline at end of file diff --git a/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area.go b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area.go new file mode 100644 index 00000000..91872524 --- /dev/null +++ b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area.go @@ -0,0 +1,31 @@ +package leetcode + +func largestTriangleArea(points [][]int) float64 { + maxArea, n := 0.0, len(points) + for i := 0; i < n; i++ { + for j := i + 1; j < n; j++ { + for k := j + 1; k < n; k++ { + maxArea = max(maxArea, area(points[i], points[j], points[k])) + } + } + } + return maxArea +} + +func area(p1, p2, p3 []int) float64 { + return abs(p1[0]*p2[1]+p2[0]*p3[1]+p3[0]*p1[1]-p1[0]*p3[1]-p2[0]*p1[1]-p3[0]*p2[1]) / 2 +} + +func abs(num int) float64 { + if num < 0 { + num = -num + } + return float64(num) +} + +func max(a, b float64) float64 { + if a > b { + return a + } + return b +} diff --git a/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area_test.go b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area_test.go new file mode 100644 index 00000000..8e461ca2 --- /dev/null +++ b/leetcode/0812.Largest-Triangle-Area/812. Largest Triangle Area_test.go @@ -0,0 +1,42 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question812 struct { + para812 + ans812 +} + +// para 是参数 +// one 代表第一个参数 +type para812 struct { + one [][]int +} + +// ans 是答案 +// one 代表第一个答案 +type ans812 struct { + one float64 +} + +func Test_Problem812(t *testing.T) { + + qs := []question812{ + + question812{ + para812{[][]int{[]int{0, 0}, []int{0, 1}, []int{1, 0}, []int{0, 2}, []int{2, 0}}}, + ans812{2.0}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 812------------------------\n") + + for _, q := range qs { + _, p := q.ans812, q.para812 + fmt.Printf("【input】:%v 【output】:%v\n", p, largestTriangleArea(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0812.Largest-Triangle-Area/README.md b/leetcode/0812.Largest-Triangle-Area/README.md new file mode 100644 index 00000000..426826f1 --- /dev/null +++ b/leetcode/0812.Largest-Triangle-Area/README.md @@ -0,0 +1,70 @@ +# [812. Largest Triangle Area](https://leetcode.com/problems/largest-triangle-area/) + + +## 题目 + +You have a list of points in the plane. Return the area of the largest triangle that can be formed by any 3 of the points. + +``` +Example: +Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]] +Output: 2 +Explanation: +The five points are show in the figure below. The red triangle is the largest. +``` + +![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png) + +**Notes**: + +- `3 <= points.length <= 50`. +- No points will be duplicated. +- `-50 <= points[i][j] <= 50`. +- Answers within `10^-6` of the true value will be accepted as correct. + +## 题目大意 + +给定包含多个点的集合，从其中取三个点组成三角形，返回能组成的最大三角形的面积。 + +## 解题思路 + +- 给出一组点的坐标，要求找出能组成三角形面积最大的点集合，输出这个最大面积。 +- 数学题。按照数学定义，分别计算这些能构成三角形的点形成的三角形面积，最终输出最大面积即可。 + +## 代码 + +```go + +package leetcode + +func largestTriangleArea(points [][]int) float64 { + maxArea, n := 0.0, len(points) + for i := 0; i < n; i++ { + for j := i + 1; j < n; j++ { + for k := j + 1; k < n; k++ { + maxArea = max(maxArea, area(points[i], points[j], points[k])) + } + } + } + return maxArea +} + +func area(p1, p2, p3 []int) float64 { + return abs(p1[0]*p2[1]+p2[0]*p3[1]+p3[0]*p1[1]-p1[0]*p3[1]-p2[0]*p1[1]-p3[0]*p2[1]) / 2 +} + +func abs(num int) float64 { + if num < 0 { + num = -num + } + return float64(num) +} + +func max(a, b float64) float64 { + if a > b { + return a + } + return b +} + +``` \ No newline at end of file diff --git a/leetcode/0832.Flipping-an-Image/832. Flipping an Image.go b/leetcode/0832.Flipping-an-Image/832. Flipping an Image.go new file mode 100644 index 00000000..6abccf66 --- /dev/null +++ b/leetcode/0832.Flipping-an-Image/832. Flipping an Image.go @@ -0,0 +1,13 @@ +package leetcode + +func flipAndInvertImage(A [][]int) [][]int { + for i := 0; i < len(A); i++ { + for a, b := 0, len(A[i])-1; a < b; a, b = a+1, b-1 { + A[i][a], A[i][b] = A[i][b], A[i][a] + } + for a := 0; a < len(A[i]); a++ { + A[i][a] = (A[i][a] + 1) % 2 + } + } + return A +} diff --git a/leetcode/0832.Flipping-an-Image/832. Flipping an Image_test.go b/leetcode/0832.Flipping-an-Image/832. Flipping an Image_test.go new file mode 100644 index 00000000..de639ef7 --- /dev/null +++ b/leetcode/0832.Flipping-an-Image/832. Flipping an Image_test.go @@ -0,0 +1,52 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question832 struct { + para832 + ans832 +} + +// para 是参数 +// one 代表第一个参数 +type para832 struct { + A [][]int +} + +// ans 是答案 +// one 代表第一个答案 +type ans832 struct { + one [][]int +} + +func Test_Problem832(t *testing.T) { + + qs := []question832{ + + question832{ + para832{[][]int{[]int{1, 1, 0}, []int{1, 0, 1}, []int{0, 0, 0}}}, + ans832{[][]int{[]int{1, 0, 0}, []int{0, 1, 0}, []int{1, 1, 1}}}, + }, + + question832{ + para832{[][]int{[]int{1, 1, 0, 0}, []int{1, 0, 0, 1}, []int{0, 1, 1, 1}, []int{1, 0, 1, 0}}}, + ans832{[][]int{[]int{1, 1, 0, 0}, []int{0, 1, 1, 0}, []int{0, 0, 0, 1}, []int{1, 0, 1, 0}}}, + }, + + question832{ + para832{[][]int{[]int{1, 1, 1}, []int{1, 1, 1}, []int{0, 0, 0}}}, + ans832{[][]int{[]int{0, 0, 0}, []int{0, 0, 0}, []int{1, 1, 1}}}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 832------------------------\n") + + for _, q := range qs { + _, p := q.ans832, q.para832 + fmt.Printf("【input】:%v 【output】:%v\n", p, flipAndInvertImage(p.A)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0832.Flipping-an-Image/README.md b/leetcode/0832.Flipping-an-Image/README.md new file mode 100644 index 00000000..81b8ccc9 --- /dev/null +++ b/leetcode/0832.Flipping-an-Image/README.md @@ -0,0 +1,63 @@ +# [832. Flipping an Image](https://leetcode.com/problems/flipping-an-image/) + + +## 题目 + +Given a binary matrix `A`, we want to flip the image horizontally, then invert it, and return the resulting image. + +To flip an image horizontally means that each row of the image is reversed. For example, flipping `[1, 1, 0]` horizontally results in `[0, 1, 1]`. + +To invert an image means that each `0` is replaced by `1`, and each `1` is replaced by `0`. For example, inverting `[0, 1, 1]` results in `[1, 0, 0]`. + +**Example 1**: + +``` +Input: [[1,1,0],[1,0,1],[0,0,0]] +Output: [[1,0,0],[0,1,0],[1,1,1]] +Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]]. +Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]] +``` + +**Example 2**: + +``` +Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]] +Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]] +Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]]. +Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]] +``` + +**Notes**: + +- `1 <= A.length = A[0].length <= 20` +- `0 <= A[i][j] <= 1` + +## 题目大意 + +给定一个二进制矩阵 A，我们想先水平翻转图像，然后反转图像并返回结果。水平翻转图片就是将图片的每一行都进行翻转，即逆序。例如，水平翻转 [1, 1, 0] 的结果是 [0, 1, 1]。反转图片的意思是图片中的 0 全部被 1 替换， 1 全部被 0 替换。例如，反转 [0, 1, 1] 的结果是 [1, 0, 0]。 + + +## 解题思路 + +- 给定一个二进制矩阵，要求先水平翻转，然后再反转( 1→0 , 0→1 )。 +- 简单题，按照题意先水平翻转，再反转即可。 + +## 代码 + +```go + +package leetcode + +func flipAndInvertImage(A [][]int) [][]int { + for i := 0; i < len(A); i++ { + for a, b := 0, len(A[i])-1; a < b; a, b = a+1, b-1 { + A[i][a], A[i][b] = A[i][b], A[i][a] + } + for a := 0; a < len(A[i]); a++ { + A[i][a] = (A[i][a] + 1) % 2 + } + } + return A +} + +``` \ No newline at end of file diff --git a/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes.go b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes.go new file mode 100644 index 00000000..c52d80d5 --- /dev/null +++ b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes.go @@ -0,0 +1,41 @@ +package leetcode + +func surfaceArea(grid [][]int) int { + area := 0 + for i := 0; i < len(grid); i++ { + for j := 0; j < len(grid[0]); j++ { + if grid[i][j] == 0 { + continue + } + area += grid[i][j]*4 + 2 + // up + if i > 0 { + m := min(grid[i][j], grid[i-1][j]) + area -= m + } + // down + if i < len(grid)-1 { + m := min(grid[i][j], grid[i+1][j]) + area -= m + } + // left + if j > 0 { + m := min(grid[i][j], grid[i][j-1]) + area -= m + } + // right + if j < len(grid[i])-1 { + m := min(grid[i][j], grid[i][j+1]) + area -= m + } + } + } + return area +} + +func min(a, b int) int { + if a > b { + return b + } + return a +} diff --git a/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes_test.go b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes_test.go new file mode 100644 index 00000000..e487bf96 --- /dev/null +++ b/leetcode/0892.Surface-Area-of-3D-Shapes/892. Surface Area of 3D Shapes_test.go @@ -0,0 +1,62 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question892 struct { + para892 + ans892 +} + +// para 是参数 +// one 代表第一个参数 +type para892 struct { + one [][]int +} + +// ans 是答案 +// one 代表第一个答案 +type ans892 struct { + one int +} + +func Test_Problem892(t *testing.T) { + + qs := []question892{ + + question892{ + para892{[][]int{[]int{2}}}, + ans892{10}, + }, + + question892{ + para892{[][]int{[]int{1, 2}, []int{3, 4}}}, + ans892{34}, + }, + + question892{ + para892{[][]int{[]int{1, 0}, []int{0, 2}}}, + ans892{16}, + }, + + question892{ + para892{[][]int{[]int{1, 1, 1}, []int{1, 0, 1}, []int{1, 1, 1}}}, + ans892{32}, + }, + + question892{ + para892{[][]int{[]int{2, 2, 2}, []int{2, 1, 2}, []int{2, 2, 2}}}, + ans892{46}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 892------------------------\n") + + for _, q := range qs { + _, p := q.ans892, q.para892 + fmt.Printf("【input】:%v 【output】:%v\n", p, surfaceArea(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0892.Surface-Area-of-3D-Shapes/README.md b/leetcode/0892.Surface-Area-of-3D-Shapes/README.md new file mode 100644 index 00000000..bcb6ad51 --- /dev/null +++ b/leetcode/0892.Surface-Area-of-3D-Shapes/README.md @@ -0,0 +1,108 @@ +# [892. Surface Area of 3D Shapes](https://leetcode.com/problems/surface-area-of-3d-shapes/) + + +## 题目 + +On a `N * N` grid, we place some `1 * 1 * 1` cubes. + +Each value `v = grid[i][j]` represents a tower of `v` cubes placed on top of grid cell `(i, j)`. + +Return the total surface area of the resulting shapes. + +**Example 1**: + +``` +Input: [[2]] +Output: 10 +``` + +**Example 2**: + +``` +Input: [[1,2],[3,4]] +Output: 34 +``` + +**Example 3**: + +``` +Input: [[1,0],[0,2]] +Output: 16 +``` + +**Example 4**: + +``` +Input: [[1,1,1],[1,0,1],[1,1,1]] +Output: 32 +``` + +**Example 5**: + +``` +Input: [[2,2,2],[2,1,2],[2,2,2]] +Output: 46 +``` + +**Note**: + +- `1 <= N <= 50` +- `0 <= grid[i][j] <= 50` + +## 题目大意 + +在 N * N 的网格上，我们放置一些 1 * 1 * 1  的立方体。每个值 v = grid[i][j] 表示 v 个正方体叠放在对应单元格 (i, j) 上。请你返回最终形体的表面积。 + + +## 解题思路 + +- 给定一个网格数组，数组里面装的是立方体叠放在所在的单元格，求最终这些叠放的立方体的表面积。 +- 简单题。按照题目意思，找到叠放时，重叠的面，然后用总表面积减去这些重叠的面积即为最终答案。 + +## 代码 + +```go + +package leetcode + +func surfaceArea(grid [][]int) int { + area := 0 + for i := 0; i < len(grid); i++ { + for j := 0; j < len(grid[0]); j++ { + if grid[i][j] == 0 { + continue + } + area += grid[i][j]*4 + 2 + // up + if i > 0 { + m := min(grid[i][j], grid[i-1][j]) + area -= m + } + // down + if i < len(grid)-1 { + m := min(grid[i][j], grid[i+1][j]) + area -= m + } + // left + if j > 0 { + m := min(grid[i][j], grid[i][j-1]) + area -= m + } + // right + if j < len(grid[i])-1 { + m := min(grid[i][j], grid[i][j+1]) + area -= m + } + } + } + return area +} + +func min(a, b int) int { + if a > b { + return b + } + return a +} + +``` \ No newline at end of file diff --git a/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits.go b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits.go new file mode 100644 index 00000000..12919863 --- /dev/null +++ b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits.go @@ -0,0 +1,33 @@ +package leetcode + +import "fmt" + +func largestTimeFromDigits(A []int) string { + flag, res := false, 0 + for i := 0; i < 4; i++ { + for j := 0; j < 4; j++ { + if i == j { + continue + } + for k := 0; k < 4; k++ { + if i == k || j == k { + continue + } + l := 6 - i - j - k + hour := A[i]*10 + A[j] + min := A[k]*10 + A[l] + if hour < 24 && min < 60 { + if hour*60+min >= res { + res = hour*60 + min + flag = true + } + } + } + } + } + if flag { + return fmt.Sprintf("%02d:%02d", res/60, res%60) + } else { + return "" + } +} diff --git a/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits_test.go b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits_test.go new file mode 100644 index 00000000..782f779b --- /dev/null +++ b/leetcode/0949.Largest-Time-for-Given-Digits/949. Largest Time for Given Digits_test.go @@ -0,0 +1,46 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question949 struct { + para949 + ans949 +} + +// para 是参数 +// one 代表第一个参数 +type para949 struct { + one []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans949 struct { + one string +} + +func Test_Problem949(t *testing.T) { + + qs := []question949{ + question949{ + para949{[]int{1, 2, 3, 4}}, + ans949{"23:41"}, + }, + + question949{ + para949{[]int{5, 5, 5, 5}}, + ans949{""}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 949------------------------\n") + + for _, q := range qs { + _, p := q.ans949, q.para949 + fmt.Printf("【input】:%v 【output】:%v\n", p, largestTimeFromDigits(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/0949.Largest-Time-for-Given-Digits/README.md b/leetcode/0949.Largest-Time-for-Given-Digits/README.md new file mode 100644 index 00000000..02f0bf43 --- /dev/null +++ b/leetcode/0949.Largest-Time-for-Given-Digits/README.md @@ -0,0 +1,78 @@ +# [949. Largest Time for Given Digits](https://leetcode.com/problems/largest-time-for-given-digits/) + + +## 题目 + +Given an array of 4 digits, return the largest 24 hour time that can be made. + +The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight. + +Return the answer as a string of length 5. If no valid time can be made, return an empty string. + +**Example 1**: + +``` +Input: [1,2,3,4] +Output: "23:41" +``` + +**Example 2**: + +``` +Input: [5,5,5,5] +Output: "" +``` + +**Note**: + +1. `A.length == 4` +2. `0 <= A[i] <= 9` + +## 题目大意 + +给定一个由 4 位数字组成的数组，返回可以设置的符合 24 小时制的最大时间。最小的 24 小时制时间是 00:00，而最大的是 23:59。从 00:00 （午夜）开始算起，过得越久，时间越大。以长度为 5 的字符串返回答案。如果不能确定有效时间，则返回空字符串。 + +## 解题思路 + +- 给出 4 个数字，要求返回一个字符串，代表由这 4 个数字能组成的最大 24 小时制的时间。 +- 简单题，这一题直接暴力枚举就可以了。依次检查给出的 4 个数字每个排列组合是否是时间合法的。例如检查 10 * A[i] + A[j] 是不是小于 24， 10 * A[k] + A[l] 是不是小于 60。如果合法且比目前存在的最大时间更大，就更新这个最大时间。 + +## 代码 + +```go + +package leetcode + +import "fmt" + +func largestTimeFromDigits(A []int) string { + flag, res := false, 0 + for i := 0; i < 4; i++ { + for j := 0; j < 4; j++ { + if i == j { + continue + } + for k := 0; k < 4; k++ { + if i == k || j == k { + continue + } + l := 6 - i - j - k + hour := A[i]*10 + A[j] + min := A[k]*10 + A[l] + if hour < 24 && min < 60 { + if hour*60+min >= res { + res = hour*60 + min + flag = true + } + } + } + } + } + if flag { + return fmt.Sprintf("%02d:%02d", res/60, res%60) + } else { + return "" + } +} + +``` \ No newline at end of file diff --git a/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang.go b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang.go new file mode 100644 index 00000000..825ae850 --- /dev/null +++ b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang.go @@ -0,0 +1,5 @@ +package leetcode + +func isBoomerang(points [][]int) bool { + return (points[0][0]-points[1][0])*(points[0][1]-points[2][1]) != (points[0][0]-points[2][0])*(points[0][1]-points[1][1]) +} diff --git a/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang_test.go b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang_test.go new file mode 100644 index 00000000..db7b4c52 --- /dev/null +++ b/leetcode/1037.Valid-Boomerang/1037. Valid Boomerang_test.go @@ -0,0 +1,46 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1037 struct { + para1037 + ans1037 +} + +// para 是参数 +// one 代表第一个参数 +type para1037 struct { + one [][]int +} + +// ans 是答案 +// one 代表第一个答案 +type ans1037 struct { + one bool +} + +func Test_Problem1037(t *testing.T) { + + qs := []question1037{ + question1037{ + para1037{[][]int{[]int{1, 2}, []int{2, 3}, []int{3, 2}}}, + ans1037{true}, + }, + + question1037{ + para1037{[][]int{[]int{1, 1}, []int{2, 2}, []int{3, 3}}}, + ans1037{false}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1037------------------------\n") + + for _, q := range qs { + _, p := q.ans1037, q.para1037 + fmt.Printf("【input】:%v 【output】:%v\n", p, isBoomerang(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/1037.Valid-Boomerang/README.md b/leetcode/1037.Valid-Boomerang/README.md new file mode 100644 index 00000000..89a4000d --- /dev/null +++ b/leetcode/1037.Valid-Boomerang/README.md @@ -0,0 +1,49 @@ +# [1037. Valid Boomerang](https://leetcode.com/problems/valid-boomerang/) + + +## 题目 + +A *boomerang* is a set of 3 points that are all distinct and **not** in a straight line. + +Given a list of three points in the plane, return whether these points are a boomerang. + +**Example 1**: + +``` +Input: [[1,1],[2,3],[3,2]] +Output: true +``` + +**Example 2**: + +``` +Input: [[1,1],[2,2],[3,3]] +Output: false +``` + +**Note**: + +1. `points.length == 3` +2. `points[i].length == 2` +3. `0 <= points[i][j] <= 100` + +## 题目大意 + +回旋镖定义为一组三个点，这些点各不相同且不在一条直线上。给出平面上三个点组成的列表，判断这些点是否可以构成回旋镖。 + +## 解题思路 + +- 判断给出的 3 组点能否满足回旋镖。 +- 简单题。判断 3 个点组成的 2 条直线的斜率是否相等。由于斜率的计算是除法，还可能遇到分母为 0 的情况，那么可以转换成乘法，交叉相乘再判断是否相等，就可以省去判断分母为 0 的情况了，代码也简洁成一行了。 + +## 代码 + +```go + +package leetcode + +func isBoomerang(points [][]int) bool { + return (points[0][0]-points[1][0])*(points[0][1]-points[2][1]) != (points[0][0]-points[2][0])*(points[0][1]-points[1][1]) +} + +``` \ No newline at end of file diff --git a/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List.go b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List.go new file mode 100644 index 00000000..9e7e7496 --- /dev/null +++ b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List.go @@ -0,0 +1,11 @@ +package leetcode + +func decompressRLElist(nums []int) []int { + res := []int{} + for i := 0; i < len(nums); i += 2 { + for j := 0; j < nums[i]; j++ { + res = append(res, nums[i+1]) + } + } + return res +} diff --git a/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List_test.go b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List_test.go new file mode 100644 index 00000000..18608311 --- /dev/null +++ b/leetcode/1313.Decompress-Run-Length-Encoded-List/1313. Decompress Run-Length Encoded List_test.go @@ -0,0 +1,53 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1313 struct { + para1313 + ans1313 +} + +// para 是参数 +// one 代表第一个参数 +type para1313 struct { + nums []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans1313 struct { + one []int +} + +func Test_Problem1313(t *testing.T) { + + qs := []question1313{ + + question1313{ + para1313{[]int{1, 2, 3, 4}}, + ans1313{[]int{2, 4, 4, 4}}, + }, + + question1313{ + para1313{[]int{1, 1, 2, 3}}, + ans1313{[]int{1, 3, 3}}, + }, + + question1313{ + para1313{[]int{}}, + ans1313{[]int{}}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1313------------------------\n") + + for _, q := range qs { + _, p := q.ans1313, q.para1313 + fmt.Printf("【input】:%v ", p) + fmt.Printf("【output】:%v \n", decompressRLElist(p.nums)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/1313.Decompress-Run-Length-Encoded-List/README.md b/leetcode/1313.Decompress-Run-Length-Encoded-List/README.md new file mode 100644 index 00000000..645d9057 --- /dev/null +++ b/leetcode/1313.Decompress-Run-Length-Encoded-List/README.md @@ -0,0 +1,60 @@ +# [1313. Decompress Run-Length Encoded List](https://leetcode.com/problems/decompress-run-length-encoded-list/) + + +## 题目 + +We are given a list `nums` of integers representing a list compressed with run-length encoding. + +Consider each adjacent pair of elements `[freq, val] = [nums[2*i], nums[2*i+1]]` (with `i >= 0`). For each such pair, there are `freq` elements with value `val` concatenated in a sublist. Concatenate all the sublists from left to right to generate the decompressed list. + +Return the decompressed list. + +**Example 1**: + +``` +Input: nums = [1,2,3,4] +Output: [2,4,4,4] +Explanation: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2]. +The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4]. +At the end the concatenation [2] + [4,4,4] is [2,4,4,4]. +``` + +**Example 2**: + +``` +Input: nums = [1,1,2,3] +Output: [1,3,3] +``` + +**Constraints**: + +- `2 <= nums.length <= 100` +- `nums.length % 2 == 0` +- `1 <= nums[i] <= 100` + +## 题目大意 + +给你一个以行程长度编码压缩的整数列表 nums 。考虑每对相邻的两个元素 [freq, val] = [nums[2*i], nums[2*i+1]] （其中 i >= 0 ），每一对都表示解压后子列表中有 freq 个值为 val 的元素，你需要从左到右连接所有子列表以生成解压后的列表。请你返回解压后的列表。 + +## 解题思路 + +- 给定一个带编码长度的数组，要求解压这个数组。 +- 简单题。按照题目要求，下标从 0 开始，奇数位下标为前一个下标对应元素重复次数，那么就把这个元素 append 几次。最终输出解压后的数组即可。 + +## 代码 + +```go + +package leetcode + +func decompressRLElist(nums []int) []int { + res := []int{} + for i := 0; i < len(nums); i += 2 { + for j := 0; j < nums[i]; j++ { + res = append(res, nums[i+1]) + } + } + return res +} + +``` \ No newline at end of file diff --git a/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers.go b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers.go new file mode 100644 index 00000000..df59ca08 --- /dev/null +++ b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers.go @@ -0,0 +1,22 @@ +package leetcode + +func getNoZeroIntegers(n int) []int { + noZeroPair := []int{} + for i := 1; i <= n/2; i++ { + if isNoZero(i) && isNoZero(n-i) { + noZeroPair = append(noZeroPair, []int{i, n - i}...) + break + } + } + return noZeroPair +} + +func isNoZero(n int) bool { + for n != 0 { + if n%10 == 0 { + return false + } + n /= 10 + } + return true +} diff --git a/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers_test.go b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers_test.go new file mode 100644 index 00000000..a4283613 --- /dev/null +++ b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/1317. Convert Integer to the Sum of Two No-Zero Integers_test.go @@ -0,0 +1,82 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1317 struct { + para1317 + ans1317 +} + +// para 是参数 +// one 代表第一个参数 +type para1317 struct { + one int +} + +// ans 是答案 +// one 代表第一个答案 +type ans1317 struct { + one []int +} + +func Test_Problem1317(t *testing.T) { + + qs := []question1317{ + + question1317{ + para1317{5}, + ans1317{[]int{1, 4}}, + }, + + question1317{ + para1317{0}, + ans1317{[]int{}}, + }, + + question1317{ + para1317{3}, + ans1317{[]int{1, 2}}, + }, + + question1317{ + para1317{1}, + ans1317{[]int{}}, + }, + + question1317{ + para1317{2}, + ans1317{[]int{1, 1}}, + }, + + question1317{ + para1317{11}, + ans1317{[]int{2, 9}}, + }, + + question1317{ + para1317{10000}, + ans1317{[]int{1, 9999}}, + }, + + question1317{ + para1317{69}, + ans1317{[]int{1, 68}}, + }, + + question1317{ + para1317{1010}, + ans1317{[]int{11, 999}}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1317------------------------\n") + + for _, q := range qs { + _, p := q.ans1317, q.para1317 + fmt.Printf("【input】:%v 【output】:%v\n", p, getNoZeroIntegers(p.one)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/README.md b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/README.md new file mode 100644 index 00000000..b7871908 --- /dev/null +++ b/leetcode/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers/README.md @@ -0,0 +1,96 @@ +# [1317. Convert Integer to the Sum of Two No-Zero Integers](https://leetcode.com/problems/convert-integer-to-the-sum-of-two-no-zero-integers/) + + +## 题目 + +Given an integer `n`. No-Zero integer is a positive integer which **doesn't contain any 0** in its decimal representation. + +Return *a list of two integers* `[A, B]` where: + +- `A` and `B` are No-Zero integers. +- `A + B = n` + +It's guarateed that there is at least one valid solution. If there are many valid solutions you can return any of them. + +**Example 1**: + +``` +Input: n = 2 +Output: [1,1] +Explanation: A = 1, B = 1. A + B = n and both A and B don't contain any 0 in their decimal representation. +``` + +**Example 2**: + +``` +Input: n = 11 +Output: [2,9] +``` + +**Example 3**: + +``` +Input: n = 10000 +Output: [1,9999] +``` + +**Example 4**: + +``` +Input: n = 69 +Output: [1,68] +``` + +**Example 5**: + +``` +Input: n = 1010 +Output: [11,999] +``` + +**Constraints**: + +- `2 <= n <= 10^4` + +## 题目大意 + +「无零整数」是十进制表示中 不含任何 0 的正整数。给你一个整数 n，请你返回一个 由两个整数组成的列表 [A, B]，满足： + +- A 和 B 都是无零整数 +- A + B = n + +题目数据保证至少有一个有效的解决方案。如果存在多个有效解决方案，你可以返回其中任意一个。 + +## 解题思路 + +- 给定一个整数 n，要求把它分解为 2 个十进制位中不含 0 的正整数且这两个正整数之和为 n。 +- 简单题。在 [1, n/2] 区间内搜索，只要有一组满足条件的解就 break。题目保证了至少有一组解，并且多组解返回任意一组即可。 + +## 代码 + +```go + +package leetcode + +func getNoZeroIntegers(n int) []int { + noZeroPair := []int{} + for i := 1; i <= n/2; i++ { + if isNoZero(i) && isNoZero(n-i) { + noZeroPair = append(noZeroPair, []int{i, n - i}...) + break + } + } + return noZeroPair +} + +func isNoZero(n int) bool { + for n != 0 { + if n%10 == 0 { + return false + } + n /= 10 + } + return true +} + +``` \ No newline at end of file diff --git a/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence.go b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence.go new file mode 100644 index 00000000..c5f13249 --- /dev/null +++ b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence.go @@ -0,0 +1,12 @@ +package leetcode + +import "strings" + +func isPrefixOfWord(sentence string, searchWord string) int { + for i, v := range strings.Split(sentence, " ") { + if strings.HasPrefix(v, searchWord) { + return i + 1 + } + } + return -1 +} diff --git a/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence_test.go b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence_test.go new file mode 100644 index 00000000..22607f7c --- /dev/null +++ b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence_test.go @@ -0,0 +1,64 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1455 struct { + para1455 + ans1455 +} + +// para 是参数 +// one 代表第一个参数 +type para1455 struct { + sentence string + searchWord string +} + +// ans 是答案 +// one 代表第一个答案 +type ans1455 struct { + one int +} + +func Test_Problem1455(t *testing.T) { + + qs := []question1455{ + + question1455{ + para1455{"i love eating burger", "burg"}, + ans1455{4}, + }, + + question1455{ + para1455{"this problem is an easy problem", "pro"}, + ans1455{2}, + }, + + question1455{ + para1455{"i am tired", "you"}, + ans1455{-1}, + }, + + question1455{ + para1455{"i use triple pillow", "pill"}, + ans1455{4}, + }, + + question1455{ + para1455{"hello from the other side", "they"}, + ans1455{-1}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1455------------------------\n") + + for _, q := range qs { + _, p := q.ans1455, q.para1455 + fmt.Printf("【input】:%v ", p) + fmt.Printf("【output】:%v \n", isPrefixOfWord(p.sentence, p.searchWord)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/README.md b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/README.md new file mode 100644 index 00000000..c8081e82 --- /dev/null +++ b/leetcode/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence/README.md @@ -0,0 +1,98 @@ +# [1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence](https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/) + + +## 题目 + +Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`. + +You have to check if `searchWord` is a prefix of any word in `sentence`. + +Return *the index of the word* in `sentence` where `searchWord` is a prefix of this word (**1-indexed**). + +If `searchWord` is a prefix of more than one word, return the index of the first word **(minimum index)**. If there is no such word return **-1**. + +A **prefix** of a string `S` is any leading contiguous substring of `S`. + +**Example 1**: + +``` +Input: sentence = "i love eating burger", searchWord = "burg" +Output: 4 +Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence. + +``` + +**Example 2**: + +``` +Input: sentence = "this problem is an easy problem", searchWord = "pro" +Output: 2 +Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index. + +``` + +**Example 3**: + +``` +Input: sentence = "i am tired", searchWord = "you" +Output: -1 +Explanation: "you" is not a prefix of any word in the sentence. + +``` + +**Example 4**: + +``` +Input: sentence = "i use triple pillow", searchWord = "pill" +Output: 4 + +``` + +**Example 5**: + +``` +Input: sentence = "hello from the other side", searchWord = "they" +Output: -1 + +``` + +**Constraints**: + +- `1 <= sentence.length <= 100` +- `1 <= searchWord.length <= 10` +- `sentence` consists of lowercase English letters and spaces. +- `searchWord` consists of lowercase English letters. + +## 题目大意 + +给你一个字符串 sentence 作为句子并指定检索词为 searchWord ，其中句子由若干用 单个空格 分隔的单词组成。请你检查检索词 searchWord 是否为句子 sentence 中任意单词的前缀。 + +- 如果 searchWord 是某一个单词的前缀，则返回句子 sentence 中该单词所对应的下标（下标从 1 开始）。 +- 如果 searchWord 是多个单词的前缀，则返回匹配的第一个单词的下标（最小下标）。 +- 如果 searchWord 不是任何单词的前缀，则返回 -1 。 + +字符串 S 的 「前缀」是 S 的任何前导连续子字符串。 + +## 解题思路 + +- 给出 2 个字符串，一个是匹配串，另外一个是句子。在句子里面查找带匹配串前缀的单词，并返回第一个匹配单词的下标。 +- 简单题。按照题意，扫描一遍句子，一次匹配即可。 + +## 代码 + +```go + +package leetcode + +import "strings" + +func isPrefixOfWord(sentence string, searchWord string) int { + for i, v := range strings.Split(sentence, " ") { + if strings.HasPrefix(v, searchWord) { + return i + 1 + } + } + return -1 +} + +``` \ No newline at end of file diff --git a/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array.go b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array.go new file mode 100644 index 00000000..a88bccb1 --- /dev/null +++ b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array.go @@ -0,0 +1,14 @@ +package leetcode + +func maxProduct(nums []int) int { + max1, max2 := 0, 0 + for _, num := range nums { + if num >= max1 { + max2 = max1 + max1 = num + } else if num <= max1 && num >= max2 { + max2 = num + } + } + return (max1 - 1) * (max2 - 1) +} diff --git a/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array_test.go b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array_test.go new file mode 100644 index 00000000..06c9821f --- /dev/null +++ b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/1464. Maximum Product of Two Elements in an Array_test.go @@ -0,0 +1,58 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1464 struct { + para1464 + ans1464 +} + +// para 是参数 +// one 代表第一个参数 +type para1464 struct { + nums []int +} + +// ans 是答案 +// one 代表第一个答案 +type ans1464 struct { + one int +} + +func Test_Problem1464(t *testing.T) { + + qs := []question1464{ + + question1464{ + para1464{[]int{3, 4, 5, 2}}, + ans1464{12}, + }, + + question1464{ + para1464{[]int{1, 5, 4, 5}}, + ans1464{16}, + }, + + question1464{ + para1464{[]int{3, 7}}, + ans1464{12}, + }, + + question1464{ + para1464{[]int{1}}, + ans1464{0}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1464------------------------\n") + + for _, q := range qs { + _, p := q.ans1464, q.para1464 + fmt.Printf("【input】:%v ", p) + fmt.Printf("【output】:%v \n", maxProduct(p.nums)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/README.md b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/README.md new file mode 100644 index 00000000..04dbf134 --- /dev/null +++ b/leetcode/1464.Maximum-Product-of-Two-Elements-in-an-Array/README.md @@ -0,0 +1,66 @@ +# [1464. Maximum Product of Two Elements in an Array](https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/) + + +## 题目 + +Given the array of integers `nums`, you will choose two different indices `i` and `j` of that array. Return the maximum value of `(nums[i]-1)*(nums[j]-1)`. + +**Example 1**: + +``` +Input: nums = [3,4,5,2] +Output: 12 +Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12. + +``` + +**Example 2**: + +``` +Input: nums = [1,5,4,5] +Output: 16 +Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16. + +``` + +**Example 3**: + +``` +Input: nums = [3,7] +Output: 12 + +``` + +**Constraints**: + +- `2 <= nums.length <= 500` +- `1 <= nums[i] <= 10^3` + +## 题目大意 + +给你一个整数数组 nums，请你选择数组的两个不同下标 i 和 j，使 (nums[i]-1)*(nums[j]-1) 取得最大值。请你计算并返回该式的最大值。 + +## 解题思路 + +- 简单题。循环一次，按照题意动态维护 2 个最大值，从而也使得 `(nums[i]-1)*(nums[j]-1)` 能取到最大值。 + +## 代码 + +```go + +package leetcode + +func maxProduct(nums []int) int { + max1, max2 := 0, 0 + for _, num := range nums { + if num >= max1 { + max2 = max1 + max1 = num + } else if num <= max1 && num >= max2 { + max2 = num + } + } + return (max1 - 1) * (max2 - 1) +} + +``` \ No newline at end of file diff --git a/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array.go b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array.go new file mode 100644 index 00000000..fef82041 --- /dev/null +++ b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array.go @@ -0,0 +1,10 @@ +package leetcode + +func shuffle(nums []int, n int) []int { + result := make([]int, 0) + for i := 0; i < n; i++ { + result = append(result, nums[i]) + result = append(result, nums[n+i]) + } + return result +} diff --git a/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array_test.go b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array_test.go new file mode 100644 index 00000000..212ecef9 --- /dev/null +++ b/leetcode/1470.Shuffle-the-Array/1470. Shuffle the Array_test.go @@ -0,0 +1,53 @@ +package leetcode + +import ( + "fmt" + "testing" +) + +type question1470 struct { + para1470 + ans1470 +} + +// para 是参数 +// one 代表第一个参数 +type para1470 struct { + nums []int + n int +} + +// ans 是答案 +// one 代表第一个答案 +type ans1470 struct { + one []int +} + +func Test_Problem1470(t *testing.T) { + + qs := []question1470{ + + question1470{ + para1470{[]int{2, 5, 1, 3, 4, 7}, 3}, + ans1470{[]int{2, 3, 5, 4, 1, 7}}, + }, + + question1470{ + para1470{[]int{1, 2, 3, 4, 4, 3, 2, 1}, 4}, + ans1470{[]int{1, 4, 2, 3, 3, 2, 4, 1}}, + }, + + question1470{ + para1470{[]int{1, 1, 2, 2}, 2}, + ans1470{[]int{1, 2, 1, 2}}, + }, + } + + fmt.Printf("------------------------Leetcode Problem 1470------------------------\n") + + for _, q := range qs { + _, p := q.ans1470, q.para1470 + fmt.Printf("【input】:%v 【output】:%v \n", p, shuffle(p.nums, p.n)) + } + fmt.Printf("\n\n\n") +} diff --git a/leetcode/1470.Shuffle-the-Array/README.md b/leetcode/1470.Shuffle-the-Array/README.md new file mode 100644 index 00000000..961d200c --- /dev/null +++ b/leetcode/1470.Shuffle-the-Array/README.md @@ -0,0 +1,64 @@ +# [1470. Shuffle the Array](https://leetcode.com/problems/shuffle-the-array/) + +## 题目 + +Given the array `nums` consisting of `2n` elements in the form `[x1,x2,...,xn,y1,y2,...,yn]`. + +*Return the array in the form* `[x1,y1,x2,y2,...,xn,yn]`. + +**Example 1**: + +``` +Input: nums = [2,5,1,3,4,7], n = 3 +Output: [2,3,5,4,1,7] +Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7]. + +``` + +**Example 2**: + +``` +Input: nums = [1,2,3,4,4,3,2,1], n = 4 +Output: [1,4,2,3,3,2,4,1] + +``` + +**Example 3**: + +``` +Input: nums = [1,1,2,2], n = 2 +Output: [1,2,1,2] + +``` + +**Constraints**: + +- `1 <= n <= 500` +- `nums.length == 2n` +- `1 <= nums[i] <= 10^3` + +## 题目大意 + +给你一个数组 nums ，数组中有 2n 个元素，按 [x1,x2,...,xn,y1,y2,...,yn] 的格式排列。请你将数组按 [x1,y1,x2,y2,...,xn,yn] 格式重新排列，返回重排后的数组。 + +## 解题思路 + +- 给定一个 2n 的数组，把后 n 个元素插空放到前 n 个元素里面。输出最终完成的数组。 +- 简单题，按照题意插空即可。 + +## 代码 + +```go + +package leetcode + +func shuffle(nums []int, n int) []int { + result := make([]int, 0) + for i := 0; i < n; i++ { + result = append(result, nums[i]) + result = append(result, nums[n+i]) + } + return result +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0009.Palindrome-Number.md b/website/content/ChapterFour/0009.Palindrome-Number.md new file mode 100644 index 00000000..d01fc716 --- /dev/null +++ b/website/content/ChapterFour/0009.Palindrome-Number.md @@ -0,0 +1,69 @@ +# [9. Palindrome Number](https://leetcode.com/problems/palindrome-number/) + + +## 题目 + +Determine whether an integer is a palindrome. An integer is a palindrome when it reads the same backward as forward. + +**Example 1**: + +``` +Input: 121 +Output: true +``` + +**Example 2**: + +``` +Input: -121 +Output: false +Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome. +``` + +**Example 3**: + +``` +Input: 10 +Output: false +Explanation: Reads 01 from right to left. Therefore it is not a palindrome. +``` + +**Follow up**: + +Coud you solve it without converting the integer to a string? + +## 题目大意 + +判断一个整数是否是回文数。回文数是指正序（从左向右）和倒序（从右向左）读都是一样的整数。 + +## 解题思路 + +- 判断一个整数是不是回文数。 +- 简单题。注意会有负数的情况，负数，个位数，10 都不是回文数。其他的整数再按照回文的规则判断。 + +## 代码 + +```go + +package leetcode + +import "strconv" + +func isPalindrome(x int) bool { + if x < 0 { + return false + } + if x < 10 { + return true + } + s := strconv.Itoa(x) + length := len(s) + for i := 0; i <= length/2; i++ { + if s[i] != s[length-1-i] { + return false + } + } + return true +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0013.Roman-to-Integer.md b/website/content/ChapterFour/0013.Roman-to-Integer.md new file mode 100644 index 00000000..44d758eb --- /dev/null +++ b/website/content/ChapterFour/0013.Roman-to-Integer.md @@ -0,0 +1,132 @@ +# [13. Roman to Integer](https://leetcode.com/problems/roman-to-integer/) + + +## 题目 + +Roman numerals are represented by seven different symbols: `I`, `V`, `X`, `L`, `C`, `D` and `M`. + +``` +Symbol Value +I 1 +V 5 +X 10 +L 50 +C 100 +D 500 +M 1000 +``` + +For example, two is written as `II` in Roman numeral, just two one's added together. Twelve is written as, `XII`, which is simply `X` + `II`. The number twenty seven is written as `XXVII`, which is `XX` + `V` + `II`. + +Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not `IIII`. Instead, the number four is written as `IV`. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as `IX`. There are six instances where subtraction is used: + +- `I` can be placed before `V` (5) and `X` (10) to make 4 and 9. +- `X` can be placed before `L` (50) and `C` (100) to make 40 and 90. +- `C` can be placed before `D` (500) and `M` (1000) to make 400 and 900. + +Given a roman numeral, convert it to an integer. Input is guaranteed to be within the range from 1 to 3999. + +**Example 1**: + +``` +Input: "III" +Output: 3 +``` + +**Example 2**: + +``` +Input: "IV" +Output: 4 +``` + +**Example 3**: + +``` +Input: "IX" +Output: 9 +``` + +**Example 4**: + +``` +Input: "LVIII" +Output: 58 +Explanation: L = 50, V= 5, III = 3. +``` + +**Example 5**: + +``` +Input: "MCMXCIV" +Output: 1994 +Explanation: M = 1000, CM = 900, XC = 90 and IV = 4. +``` + +## 题目大意 + +罗马数字包含以下七种字符: I， V， X， L，C，D 和 M。 + +```go + +字符 数值 +I 1 +V 5 +X 10 +L 50 +C 100 +D 500 +M 1000 + +``` + +例如， 罗马数字 2 写做 II ，即为两个并列的 1。12 写做 XII ，即为 X + II 。 27 写做  XXVII, 即为 XX + V + II 。 + +通常情况下，罗马数字中小的数字在大的数字的右边。但也存在特例，例如 4 不写做 IIII，而是 IV。数字 1 在数字 5 的左边，所表示的数等于大数 5 减小数 1 得到的数值 4 。同样地，数字 9 表示为 IX。这个特殊的规则只适用于以下六种情况： + +- I 可以放在 V (5) 和 X (10) 的左边，来表示 4 和 9。 +- X 可以放在 L (50) 和 C (100) 的左边，来表示 40 和 90。  +- C 可以放在 D (500) 和 M (1000) 的左边，来表示 400 和 900。 + +给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。 + +## 解题思路 + +- 给定一个罗马数字，将其转换成整数。输入确保在 1 到 3999 的范围内。 +- 简单题。按照题目中罗马数字的字符数值，计算出对应罗马数字的十进制数即可。 + +## 代码 + +```go + +package leetcode + +var roman = map[string]int{ + "I": 1, + "V": 5, + "X": 10, + "L": 50, + "C": 100, + "D": 500, + "M": 1000, +} + +func romanToInt(s string) int { + if s == "" { + return 0 + } + num, lastint, total := 0, 0, 0 + for i := 0; i < len(s); i++ { + char := s[len(s)-(i+1) : len(s)-i] + num = roman[char] + if num < lastint { + total = total - num + } else { + total = total + num + } + lastint = num + } + return total +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0067.Add-Binary.md b/website/content/ChapterFour/0067.Add-Binary.md new file mode 100644 index 00000000..91d977e5 --- /dev/null +++ b/website/content/ChapterFour/0067.Add-Binary.md @@ -0,0 +1,76 @@ +# [67. Add Binary](https://leetcode.com/problems/add-binary/) + + +## 题目 + +Given two binary strings, return their sum (also a binary string). + +The input strings are both **non-empty** and contains only characters `1` or `0`. + +**Example 1**: + +``` +Input: a = "11", b = "1" +Output: "100" +``` + +**Example 2**: + +``` +Input: a = "1010", b = "1011" +Output: "10101" +``` + +## 题目大意 + +给你两个二进制字符串，返回它们的和（用二进制表示）。输入为 非空 字符串且只包含数字 1 和 0。 + +## 解题思路 + +- 要求输出 2 个二进制数的和，结果也用二进制表示。 +- 简单题。按照二进制的加法规则做加法即可。 + +## 代码 + +```go + +package leetcode + +import ( + "strconv" + "strings" +) + +func addBinary(a string, b string) string { + if len(b) > len(a) { + a, b = b, a + } + + res := make([]string, len(a)+1) + i, j, k, c := len(a)-1, len(b)-1, len(a), 0 + for i >= 0 && j >= 0 { + ai, _ := strconv.Atoi(string(a[i])) + bj, _ := strconv.Atoi(string(b[j])) + res[k] = strconv.Itoa((ai + bj + c) % 2) + c = (ai + bj + c) / 2 + i-- + j-- + k-- + } + + for i >= 0 { + ai, _ := strconv.Atoi(string(a[i])) + res[k] = strconv.Itoa((ai + c) % 2) + c = (ai + c) / 2 + i-- + k-- + } + + if c > 0 { + res[k] = strconv.Itoa(c) + } + + return strings.Join(res, "") +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0168.Excel-Sheet-Column-Title.md b/website/content/ChapterFour/0168.Excel-Sheet-Column-Title.md new file mode 100644 index 00000000..c1ecdd58 --- /dev/null +++ b/website/content/ChapterFour/0168.Excel-Sheet-Column-Title.md @@ -0,0 +1,80 @@ +# [168. Excel Sheet Column Title](https://leetcode.com/problems/excel-sheet-column-title/) + +## 题目 + +Given a positive integer, return its corresponding column title as appear in an Excel sheet. + +For example: + +``` + 1 -> A + 2 -> B + 3 -> C + ... + 26 -> Z + 27 -> AA + 28 -> AB + ... +``` + +**Example 1**: + +``` +Input: 1 +Output: "A" +``` + +**Example 2**: + +``` +Input: 28 +Output: "AB" +``` + +**Example 3**: + +``` +Input: 701 +Output: "ZY" +``` + +## 题目大意 + +给定一个正整数，返回它在 Excel 表中相对应的列名称。 + +例如， + + 1 -> A + 2 -> B + 3 -> C + ... + 26 -> Z + 27 -> AA + 28 -> AB + ... + + +## 解题思路 + +- 给定一个正整数，返回它在 Excel 表中的对应的列名称 +- 简单题。这一题就类似短除法的计算过程。以 26 进制的字母编码。按照短除法先除，然后余数逆序输出即可。 + +## 代码 + +```go + +package leetcode + +func convertToTitle(n int) string { + result := []byte{} + for n > 0 { + result = append(result, 'A'+byte((n-1)%26)) + n = (n - 1) / 26 + } + for i, j := 0, len(result)-1; i < j; i, j = i+1, j-1 { + result[i], result[j] = result[j], result[i] + } + return string(result) +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0171.Excel-Sheet-Column-Number.md b/website/content/ChapterFour/0171.Excel-Sheet-Column-Number.md new file mode 100644 index 00000000..8c9f6651 --- /dev/null +++ b/website/content/ChapterFour/0171.Excel-Sheet-Column-Number.md @@ -0,0 +1,67 @@ +# [171. Excel Sheet Column Number](https://leetcode.com/problems/excel-sheet-column-number/) + + +## 题目 + +Given a column title as appear in an Excel sheet, return its corresponding column number. + +For example: + +``` + A -> 1 + B -> 2 + C -> 3 + ... + Z -> 26 + AA -> 27 + AB -> 28 + ... +``` + +**Example 1**: + +``` +Input: "A" +Output: 1 +``` + +**Example 2**: + +``` +Input: "AB" +Output: 28 +``` + +**Example 3**: + +``` +Input: "ZY" +Output: 701 +``` + +## 题目大意 + +给定一个 Excel 表格中的列名称，返回其相应的列序号。 + + +## 解题思路 + +- 给出 Excel 中列的名称，输出其对应的列序号。 +- 简单题。这一题是第 168 题的逆序题。按照 26 进制还原成十进制即可。 + +## 代码 + +```go + +package leetcode + +func titleToNumber(s string) int { + val, res := 0, 0 + for i := 0; i < len(s); i++ { + val = int(s[i] - 'A' + 1) + res = res*26 + val + } + return res +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0258.Add-Digits.md b/website/content/ChapterFour/0258.Add-Digits.md new file mode 100644 index 00000000..8c811904 --- /dev/null +++ b/website/content/ChapterFour/0258.Add-Digits.md @@ -0,0 +1,47 @@ +# [258. Add Digits](https://leetcode.com/problems/add-digits/) + + +## 题目 + +Given a non-negative integer `num`, repeatedly add all its digits until the result has only one digit. + +**Example**: + +``` +Input: 38 +Output: 2 +Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2. + Since 2 has only one digit, return it. +``` + +**Follow up**: Could you do it without any loop/recursion in O(1) runtime? + +## 题目大意 + +给定一个非负整数 num，反复将各个位上的数字相加，直到结果为一位数。 + + +## 解题思路 + +- 给定一个非负整数，反复加各个位上的数，直到结果为一位数为止，最后输出这一位数。 +- 简单题。按照题意循环累加即可。 + +## 代码 + +```go + +package leetcode + +func addDigits(num int) int { + for num > 9 { + cur := 0 + for num != 0 { + cur += num % 10 + num /= 10 + } + num = cur + } + return num +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0453.Minimum-Moves-to-Equal-Array-Elements.md b/website/content/ChapterFour/0453.Minimum-Moves-to-Equal-Array-Elements.md new file mode 100644 index 00000000..280f22a2 --- /dev/null +++ b/website/content/ChapterFour/0453.Minimum-Moves-to-Equal-Array-Elements.md @@ -0,0 +1,51 @@ +# [453. Minimum Moves to Equal Array Elements](https://leetcode.com/problems/minimum-moves-to-equal-array-elements/) + + +## 题目 + +Given a **non-empty** integer array of size n, find the minimum number of moves required to make all array elements equal, where a move is incrementing n - 1 elements by 1. + +**Example**: + +``` +Input: +[1,2,3] + +Output: +3 + +Explanation: +Only three moves are needed (remember each move increments two elements): + +[1,2,3] => [2,3,3] => [3,4,3] => [4,4,4] +``` + +## 题目大意 + +给定一个长度为 n 的非空整数数组，找到让数组所有元素相等的最小移动次数。每次移动将会使 n - 1 个元素增加 1。 + +## 解题思路 + +- 给定一个数组，要求输出让所有元素都相等的最小步数。每移动一步都会使得 n - 1 个元素 + 1 。 +- 数学题。这道题正着思考会考虑到排序或者暴力的方法上去。反过来思考一下，使得每个元素都相同，意思让所有元素的差异变为 0 。每次移动的过程中，都有 n - 1 个元素 + 1，那么没有 + 1 的那个元素和其他 n - 1 个元素相对差异就缩小了。所以这道题让所有元素都变为相等的最少步数，即等于让所有元素相对差异减少到最小的那个数。想到这里，此题就可以优雅的解出来了。 + +## 代码 + +```go + +package leetcode + +import "math" + +func minMoves(nums []int) int { + sum, min, l := 0, math.MaxInt32, len(nums) + for _, v := range nums { + sum += v + if min > v { + min = v + } + } + return sum - min*l +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0507.Perfect-Number.md b/website/content/ChapterFour/0507.Perfect-Number.md new file mode 100644 index 00000000..12d0c764 --- /dev/null +++ b/website/content/ChapterFour/0507.Perfect-Number.md @@ -0,0 +1,64 @@ +# [507. Perfect Number](https://leetcode.com/problems/perfect-number/) + + + +## 题目 + +We define the Perfect Number is a **positive** integer that is equal to the sum of all its **positive** divisors except itself. + +Now, given an + +**integer** + +n, write a function that returns true when it is a perfect number and false when it is not. + +**Example**: + +``` +Input: 28 +Output: True +Explanation: 28 = 1 + 2 + 4 + 7 + 14 +``` + +**Note**: The input number **n** will not exceed 100,000,000. (1e8) + +## 题目大意 + +对于一个 正整数，如果它和除了它自身以外的所有正因子之和相等，我们称它为“完美数”。给定一个 整数 n， 如果他是完美数，返回 True，否则返回 False + +## 解题思路 + +- 给定一个整数，要求判断这个数是不是完美数。整数的取值范围小于 1e8 。 +- 简单题。按照题意描述，先获取这个整数的所有正因子，如果正因子的和等于原来这个数，那么它就是完美数。 +- 这一题也可以打表，1e8 以下的完美数其实并不多，就 5 个。 + +## 代码 + +```go + +package leetcode + +import "math" + +// 方法一 +func checkPerfectNumber(num int) bool { + if num <= 1 { + return false + } + sum, bound := 1, int(math.Sqrt(float64(num)))+1 + for i := 2; i < bound; i++ { + if num%i != 0 { + continue + } + corrDiv := num / i + sum += corrDiv + i + } + return sum == num +} + +// 方法二 打表 +func checkPerfectNumber_(num int) bool { + return num == 6 || num == 28 || num == 496 || num == 8128 || num == 33550336 +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0537.Complex-Number-Multiplication.md b/website/content/ChapterFour/0537.Complex-Number-Multiplication.md new file mode 100644 index 00000000..568404ba --- /dev/null +++ b/website/content/ChapterFour/0537.Complex-Number-Multiplication.md @@ -0,0 +1,73 @@ +# [537. Complex Number Multiplication](https://leetcode.com/problems/complex-number-multiplication/) + + +## 题目 + +Given two strings representing two [complex numbers](https://en.wikipedia.org/wiki/Complex_number). + +You need to return a string representing their multiplication. Note i2 = -1 according to the definition. + +**Example 1**: + +``` +Input: "1+1i", "1+1i" +Output: "0+2i" +Explanation: (1 + i) * (1 + i) = 1 + i2 + 2 * i = 2i, and you need convert it to the form of 0+2i. +``` + +**Example 2**: + +``` +Input: "1+-1i", "1+-1i" +Output: "0+-2i" +Explanation: (1 - i) * (1 - i) = 1 + i2 - 2 * i = -2i, and you need convert it to the form of 0+-2i. +``` + +**Note**: + +1. The input strings will not have extra blank. +2. The input strings will be given in the form of **a+bi**, where the integer **a** and **b** will both belong to the range of [-100, 100]. And **the output should be also in this form**. + +## 题目大意 + +给定两个表示复数的字符串。返回表示它们乘积的字符串。注意，根据定义 i^2 = -1 。 + +注意: + +- 输入字符串不包含额外的空格。 +- 输入字符串将以 a+bi 的形式给出，其中整数 a 和 b 的范围均在 [-100, 100] 之间。输出也应当符合这种形式。 + + + +## 解题思路 + +- 给定 2 个字符串，要求这两个复数的乘积，输出也是字符串格式。 +- 数学题。按照复数的运算法则，i^2 = -1，最后输出字符串结果即可。 + +## 代码 + +```go + +package leetcode + +import ( + "strconv" + "strings" +) + +func complexNumberMultiply(a string, b string) string { + realA, imagA := parse(a) + realB, imagB := parse(b) + real := realA*realB - imagA*imagB + imag := realA*imagB + realB*imagA + return strconv.Itoa(real) + "+" + strconv.Itoa(imag) + "i" +} + +func parse(s string) (int, int) { + ss := strings.Split(s, "+") + r, _ := strconv.Atoi(ss[0]) + i, _ := strconv.Atoi(ss[1][:len(ss[1])-1]) + return r, i +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0561.Array-Partition-I.md b/website/content/ChapterFour/0561.Array-Partition-I.md new file mode 100644 index 00000000..d72ea633 --- /dev/null +++ b/website/content/ChapterFour/0561.Array-Partition-I.md @@ -0,0 +1,58 @@ +# [561. Array Partition I](https://leetcode.com/problems/array-partition-i/) + + +## 题目 + +Given an array of **2n** integers, your task is to group these integers into **n** pairs of integer, say (a1, b1), (a2, b2), ..., (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible. + +**Example 1**: + +``` +Input: [1,4,3,2] + +Output: 4 +Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4). +``` + +**Note**: + +1. **n** is a positive integer, which is in the range of [1, 10000]. +2. All the integers in the array will be in the range of [-10000, 10000]. + +## 题目大意 + +给定长度为 2n 的数组, 你的任务是将这些数分成 n 对, 例如 (a1, b1), (a2, b2), ..., (an, bn) ，使得从1 到 n 的 min(ai, bi) 总和最大。 + + +## 解题思路 + +- 给定一个 2n 个数组，要求把它们分为 n 组一行，求出各组最小值的总和的最大值。 +- 由于题目给的数据范围不大，[-10000, 10000]，所以我们可以考虑用一个哈希表数组，里面存储 i - 10000 元素的频次，偏移量是 10000。这个哈希表能按递增的顺序访问数组，这样可以减少排序的耗时。题目要求求出分组以后求和的最大值，那么所有偏小的元素尽量都安排在一组里面，这样取 min 以后，对最大和影响不大。例如，(1 , 1) 这样安排在一起，min 以后就是 1 。但是如果把相差很大的两个元素安排到一起，那么较大的那个元素就“牺牲”了。例如，(1 , 10000)，取 min 以后就是 1，于是 10000 就“牺牲”了。所以需要优先考虑较小值。 +- 较小值出现的频次可能是奇数也可能是偶数。如果是偶数，那比较简单，把它们俩俩安排在一起就可以了。如果是奇数，那么它会落单一次，落单的那个需要和距离它最近的一个元素进行配对，这样对最终的和影响最小。较小值如果是奇数，那么就会影响后面元素的选择，后面元素如果是偶数，由于需要一个元素和前面的较小值配对，所以它剩下的又是奇数个。这个影响会依次传递到后面。所以用一个 flag 标记，如果当前集合中有剩余元素将被再次考虑，则此标志设置为 1。在从下一组中选择元素时，会考虑已考虑的相同额外元素。 +- 最后扫描过程中动态的维护 sum 值就可以了。 + +## 代码 + +```go + +package leetcode + +func arrayPairSum(nums []int) int { + array := [20001]int{} + for i := 0; i < len(nums); i++ { + array[nums[i]+10000]++ + } + flag, sum := true, 0 + for i := 0; i < len(array); i++ { + for array[i] > 0 { + if flag { + sum = sum + i - 10000 + } + flag = !flag + array[i]-- + } + } + return sum +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0598.Range-Addition-II.md b/website/content/ChapterFour/0598.Range-Addition-II.md new file mode 100644 index 00000000..a4606f95 --- /dev/null +++ b/website/content/ChapterFour/0598.Range-Addition-II.md @@ -0,0 +1,82 @@ +# [598. Range Addition II](https://leetcode.com/problems/range-addition-ii/) + + +## 题目 + +Given an m * n matrix **M** initialized with all **0**'s and several update operations. + +Operations are represented by a 2D array, and each operation is represented by an array with two **positive** integers **a** and **b**, which means **M[i][j]** should be **added by one** for all **0 <= i < a** and **0 <= j < b**. + +You need to count and return the number of maximum integers in the matrix after performing all the operations. + +**Example 1**: + +``` +Input: +m = 3, n = 3 +operations = [[2,2],[3,3]] +Output: 4 +Explanation: +Initially, M = +[[0, 0, 0], + [0, 0, 0], + [0, 0, 0]] + +After performing [2,2], M = +[[1, 1, 0], + [1, 1, 0], + [0, 0, 0]] + +After performing [3,3], M = +[[2, 2, 1], + [2, 2, 1], + [1, 1, 1]] + +So the maximum integer in M is 2, and there are four of it in M. So return 4. +``` + +**Note**: + +1. The range of m and n is [1,40000]. +2. The range of a is [1,m], and the range of b is [1,n]. +3. The range of operations size won't exceed 10,000. + +## 题目大意 + +给定一个初始元素全部为 0，大小为 m*n 的矩阵 M 以及在 M 上的一系列更新操作。操作用二维数组表示，其中的每个操作用一个含有两个正整数 a 和 b 的数组表示，含义是将所有符合 0 <= i < a 以及 0 <= j < b 的元素 M[i][j] 的值都增加 1。在执行给定的一系列操作后，你需要返回矩阵中含有最大整数的元素个数。 + +注意: + +- m 和 n 的范围是 [1,40000]。 +- a 的范围是 [1,m]，b 的范围是 [1,n]。 +- 操作数目不超过 10000。 + + +## 解题思路 + +- 给定一个初始都为 0 的 m * n 的矩阵，和一个操作数组。经过一系列的操作以后，最终输出矩阵中最大整数的元素个数。每次操作都使得一个矩形内的元素都 + 1 。 +- 这一题乍一看像线段树的区间覆盖问题，但是实际上很简单。如果此题是任意的矩阵，那就可能用到线段树了。这一题每个矩阵的起点都包含 [0 , 0] 这个元素，也就是说每次操作都会影响第一个元素。那么这道题就很简单了。经过 n 次操作以后，被覆盖次数最多的矩形区间，一定就是最大整数所在的区间。由于起点都是第一个元素，所以我们只用关心矩形的右下角那个坐标。右下角怎么计算呢？只用每次动态的维护一下矩阵长和宽的最小值即可。 + +## 代码 + +```go + +package leetcode + +func maxCount(m int, n int, ops [][]int) int { + minM, minN := m, n + for _, op := range ops { + minM = min(minM, op[0]) + minN = min(minN, op[1]) + } + return minM * minN +} + +func min(a, b int) int { + if a < b { + return a + } + return b +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0812.Largest-Triangle-Area.md b/website/content/ChapterFour/0812.Largest-Triangle-Area.md new file mode 100644 index 00000000..426826f1 --- /dev/null +++ b/website/content/ChapterFour/0812.Largest-Triangle-Area.md @@ -0,0 +1,70 @@ +# [812. Largest Triangle Area](https://leetcode.com/problems/largest-triangle-area/) + + +## 题目 + +You have a list of points in the plane. Return the area of the largest triangle that can be formed by any 3 of the points. + +``` +Example: +Input: points = [[0,0],[0,1],[1,0],[0,2],[2,0]] +Output: 2 +Explanation: +The five points are show in the figure below. The red triangle is the largest. +``` + +![https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png](https://s3-lc-upload.s3.amazonaws.com/uploads/2018/04/04/1027.png) + +**Notes**: + +- `3 <= points.length <= 50`. +- No points will be duplicated. +- `-50 <= points[i][j] <= 50`. +- Answers within `10^-6` of the true value will be accepted as correct. + +## 题目大意 + +给定包含多个点的集合，从其中取三个点组成三角形，返回能组成的最大三角形的面积。 + +## 解题思路 + +- 给出一组点的坐标，要求找出能组成三角形面积最大的点集合，输出这个最大面积。 +- 数学题。按照数学定义，分别计算这些能构成三角形的点形成的三角形面积，最终输出最大面积即可。 + +## 代码 + +```go + +package leetcode + +func largestTriangleArea(points [][]int) float64 { + maxArea, n := 0.0, len(points) + for i := 0; i < n; i++ { + for j := i + 1; j < n; j++ { + for k := j + 1; k < n; k++ { + maxArea = max(maxArea, area(points[i], points[j], points[k])) + } + } + } + return maxArea +} + +func area(p1, p2, p3 []int) float64 { + return abs(p1[0]*p2[1]+p2[0]*p3[1]+p3[0]*p1[1]-p1[0]*p3[1]-p2[0]*p1[1]-p3[0]*p2[1]) / 2 +} + +func abs(num int) float64 { + if num < 0 { + num = -num + } + return float64(num) +} + +func max(a, b float64) float64 { + if a > b { + return a + } + return b +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0832.Flipping-an-Image.md b/website/content/ChapterFour/0832.Flipping-an-Image.md new file mode 100644 index 00000000..81b8ccc9 --- /dev/null +++ b/website/content/ChapterFour/0832.Flipping-an-Image.md @@ -0,0 +1,63 @@ +# [832. Flipping an Image](https://leetcode.com/problems/flipping-an-image/) + + +## 题目 + +Given a binary matrix `A`, we want to flip the image horizontally, then invert it, and return the resulting image. + +To flip an image horizontally means that each row of the image is reversed. For example, flipping `[1, 1, 0]` horizontally results in `[0, 1, 1]`. + +To invert an image means that each `0` is replaced by `1`, and each `1` is replaced by `0`. For example, inverting `[0, 1, 1]` results in `[1, 0, 0]`. + +**Example 1**: + +``` +Input: [[1,1,0],[1,0,1],[0,0,0]] +Output: [[1,0,0],[0,1,0],[1,1,1]] +Explanation: First reverse each row: [[0,1,1],[1,0,1],[0,0,0]]. +Then, invert the image: [[1,0,0],[0,1,0],[1,1,1]] +``` + +**Example 2**: + +``` +Input: [[1,1,0,0],[1,0,0,1],[0,1,1,1],[1,0,1,0]] +Output: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]] +Explanation: First reverse each row: [[0,0,1,1],[1,0,0,1],[1,1,1,0],[0,1,0,1]]. +Then invert the image: [[1,1,0,0],[0,1,1,0],[0,0,0,1],[1,0,1,0]] +``` + +**Notes**: + +- `1 <= A.length = A[0].length <= 20` +- `0 <= A[i][j] <= 1` + +## 题目大意 + +给定一个二进制矩阵 A，我们想先水平翻转图像，然后反转图像并返回结果。水平翻转图片就是将图片的每一行都进行翻转，即逆序。例如，水平翻转 [1, 1, 0] 的结果是 [0, 1, 1]。反转图片的意思是图片中的 0 全部被 1 替换， 1 全部被 0 替换。例如，反转 [0, 1, 1] 的结果是 [1, 0, 0]。 + + +## 解题思路 + +- 给定一个二进制矩阵，要求先水平翻转，然后再反转( 1→0 , 0→1 )。 +- 简单题，按照题意先水平翻转，再反转即可。 + +## 代码 + +```go + +package leetcode + +func flipAndInvertImage(A [][]int) [][]int { + for i := 0; i < len(A); i++ { + for a, b := 0, len(A[i])-1; a < b; a, b = a+1, b-1 { + A[i][a], A[i][b] = A[i][b], A[i][a] + } + for a := 0; a < len(A[i]); a++ { + A[i][a] = (A[i][a] + 1) % 2 + } + } + return A +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0892.Surface-Area-of-3D-Shapes.md b/website/content/ChapterFour/0892.Surface-Area-of-3D-Shapes.md new file mode 100644 index 00000000..bcb6ad51 --- /dev/null +++ b/website/content/ChapterFour/0892.Surface-Area-of-3D-Shapes.md @@ -0,0 +1,108 @@ +# [892. Surface Area of 3D Shapes](https://leetcode.com/problems/surface-area-of-3d-shapes/) + + +## 题目 + +On a `N * N` grid, we place some `1 * 1 * 1` cubes. + +Each value `v = grid[i][j]` represents a tower of `v` cubes placed on top of grid cell `(i, j)`. + +Return the total surface area of the resulting shapes. + +**Example 1**: + +``` +Input: [[2]] +Output: 10 +``` + +**Example 2**: + +``` +Input: [[1,2],[3,4]] +Output: 34 +``` + +**Example 3**: + +``` +Input: [[1,0],[0,2]] +Output: 16 +``` + +**Example 4**: + +``` +Input: [[1,1,1],[1,0,1],[1,1,1]] +Output: 32 +``` + +**Example 5**: + +``` +Input: [[2,2,2],[2,1,2],[2,2,2]] +Output: 46 +``` + +**Note**: + +- `1 <= N <= 50` +- `0 <= grid[i][j] <= 50` + +## 题目大意 + +在 N * N 的网格上，我们放置一些 1 * 1 * 1  的立方体。每个值 v = grid[i][j] 表示 v 个正方体叠放在对应单元格 (i, j) 上。请你返回最终形体的表面积。 + + +## 解题思路 + +- 给定一个网格数组，数组里面装的是立方体叠放在所在的单元格，求最终这些叠放的立方体的表面积。 +- 简单题。按照题目意思，找到叠放时，重叠的面，然后用总表面积减去这些重叠的面积即为最终答案。 + +## 代码 + +```go + +package leetcode + +func surfaceArea(grid [][]int) int { + area := 0 + for i := 0; i < len(grid); i++ { + for j := 0; j < len(grid[0]); j++ { + if grid[i][j] == 0 { + continue + } + area += grid[i][j]*4 + 2 + // up + if i > 0 { + m := min(grid[i][j], grid[i-1][j]) + area -= m + } + // down + if i < len(grid)-1 { + m := min(grid[i][j], grid[i+1][j]) + area -= m + } + // left + if j > 0 { + m := min(grid[i][j], grid[i][j-1]) + area -= m + } + // right + if j < len(grid[i])-1 { + m := min(grid[i][j], grid[i][j+1]) + area -= m + } + } + } + return area +} + +func min(a, b int) int { + if a > b { + return b + } + return a +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/0949.Largest-Time-for-Given-Digits.md b/website/content/ChapterFour/0949.Largest-Time-for-Given-Digits.md new file mode 100644 index 00000000..02f0bf43 --- /dev/null +++ b/website/content/ChapterFour/0949.Largest-Time-for-Given-Digits.md @@ -0,0 +1,78 @@ +# [949. Largest Time for Given Digits](https://leetcode.com/problems/largest-time-for-given-digits/) + + +## 题目 + +Given an array of 4 digits, return the largest 24 hour time that can be made. + +The smallest 24 hour time is 00:00, and the largest is 23:59. Starting from 00:00, a time is larger if more time has elapsed since midnight. + +Return the answer as a string of length 5. If no valid time can be made, return an empty string. + +**Example 1**: + +``` +Input: [1,2,3,4] +Output: "23:41" +``` + +**Example 2**: + +``` +Input: [5,5,5,5] +Output: "" +``` + +**Note**: + +1. `A.length == 4` +2. `0 <= A[i] <= 9` + +## 题目大意 + +给定一个由 4 位数字组成的数组，返回可以设置的符合 24 小时制的最大时间。最小的 24 小时制时间是 00:00，而最大的是 23:59。从 00:00 （午夜）开始算起，过得越久，时间越大。以长度为 5 的字符串返回答案。如果不能确定有效时间，则返回空字符串。 + +## 解题思路 + +- 给出 4 个数字，要求返回一个字符串，代表由这 4 个数字能组成的最大 24 小时制的时间。 +- 简单题，这一题直接暴力枚举就可以了。依次检查给出的 4 个数字每个排列组合是否是时间合法的。例如检查 10 * A[i] + A[j] 是不是小于 24， 10 * A[k] + A[l] 是不是小于 60。如果合法且比目前存在的最大时间更大，就更新这个最大时间。 + +## 代码 + +```go + +package leetcode + +import "fmt" + +func largestTimeFromDigits(A []int) string { + flag, res := false, 0 + for i := 0; i < 4; i++ { + for j := 0; j < 4; j++ { + if i == j { + continue + } + for k := 0; k < 4; k++ { + if i == k || j == k { + continue + } + l := 6 - i - j - k + hour := A[i]*10 + A[j] + min := A[k]*10 + A[l] + if hour < 24 && min < 60 { + if hour*60+min >= res { + res = hour*60 + min + flag = true + } + } + } + } + } + if flag { + return fmt.Sprintf("%02d:%02d", res/60, res%60) + } else { + return "" + } +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/1037.Valid-Boomerang.md b/website/content/ChapterFour/1037.Valid-Boomerang.md new file mode 100644 index 00000000..89a4000d --- /dev/null +++ b/website/content/ChapterFour/1037.Valid-Boomerang.md @@ -0,0 +1,49 @@ +# [1037. Valid Boomerang](https://leetcode.com/problems/valid-boomerang/) + + +## 题目 + +A *boomerang* is a set of 3 points that are all distinct and **not** in a straight line. + +Given a list of three points in the plane, return whether these points are a boomerang. + +**Example 1**: + +``` +Input: [[1,1],[2,3],[3,2]] +Output: true +``` + +**Example 2**: + +``` +Input: [[1,1],[2,2],[3,3]] +Output: false +``` + +**Note**: + +1. `points.length == 3` +2. `points[i].length == 2` +3. `0 <= points[i][j] <= 100` + +## 题目大意 + +回旋镖定义为一组三个点，这些点各不相同且不在一条直线上。给出平面上三个点组成的列表，判断这些点是否可以构成回旋镖。 + +## 解题思路 + +- 判断给出的 3 组点能否满足回旋镖。 +- 简单题。判断 3 个点组成的 2 条直线的斜率是否相等。由于斜率的计算是除法，还可能遇到分母为 0 的情况，那么可以转换成乘法，交叉相乘再判断是否相等，就可以省去判断分母为 0 的情况了，代码也简洁成一行了。 + +## 代码 + +```go + +package leetcode + +func isBoomerang(points [][]int) bool { + return (points[0][0]-points[1][0])*(points[0][1]-points[2][1]) != (points[0][0]-points[2][0])*(points[0][1]-points[1][1]) +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/1313.Decompress-Run-Length-Encoded-List.md b/website/content/ChapterFour/1313.Decompress-Run-Length-Encoded-List.md new file mode 100644 index 00000000..645d9057 --- /dev/null +++ b/website/content/ChapterFour/1313.Decompress-Run-Length-Encoded-List.md @@ -0,0 +1,60 @@ +# [1313. Decompress Run-Length Encoded List](https://leetcode.com/problems/decompress-run-length-encoded-list/) + + +## 题目 + +We are given a list `nums` of integers representing a list compressed with run-length encoding. + +Consider each adjacent pair of elements `[freq, val] = [nums[2*i], nums[2*i+1]]` (with `i >= 0`). For each such pair, there are `freq` elements with value `val` concatenated in a sublist. Concatenate all the sublists from left to right to generate the decompressed list. + +Return the decompressed list. + +**Example 1**: + +``` +Input: nums = [1,2,3,4] +Output: [2,4,4,4] +Explanation: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2]. +The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4]. +At the end the concatenation [2] + [4,4,4] is [2,4,4,4]. +``` + +**Example 2**: + +``` +Input: nums = [1,1,2,3] +Output: [1,3,3] +``` + +**Constraints**: + +- `2 <= nums.length <= 100` +- `nums.length % 2 == 0` +- `1 <= nums[i] <= 100` + +## 题目大意 + +给你一个以行程长度编码压缩的整数列表 nums 。考虑每对相邻的两个元素 [freq, val] = [nums[2*i], nums[2*i+1]] （其中 i >= 0 ），每一对都表示解压后子列表中有 freq 个值为 val 的元素，你需要从左到右连接所有子列表以生成解压后的列表。请你返回解压后的列表。 + +## 解题思路 + +- 给定一个带编码长度的数组，要求解压这个数组。 +- 简单题。按照题目要求，下标从 0 开始，奇数位下标为前一个下标对应元素重复次数，那么就把这个元素 append 几次。最终输出解压后的数组即可。 + +## 代码 + +```go + +package leetcode + +func decompressRLElist(nums []int) []int { + res := []int{} + for i := 0; i < len(nums); i += 2 { + for j := 0; j < nums[i]; j++ { + res = append(res, nums[i+1]) + } + } + return res +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers.md b/website/content/ChapterFour/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers.md new file mode 100644 index 00000000..b7871908 --- /dev/null +++ b/website/content/ChapterFour/1317.Convert-Integer-to-the-Sum-of-Two-No-Zero-Integers.md @@ -0,0 +1,96 @@ +# [1317. Convert Integer to the Sum of Two No-Zero Integers](https://leetcode.com/problems/convert-integer-to-the-sum-of-two-no-zero-integers/) + + +## 题目 + +Given an integer `n`. No-Zero integer is a positive integer which **doesn't contain any 0** in its decimal representation. + +Return *a list of two integers* `[A, B]` where: + +- `A` and `B` are No-Zero integers. +- `A + B = n` + +It's guarateed that there is at least one valid solution. If there are many valid solutions you can return any of them. + +**Example 1**: + +``` +Input: n = 2 +Output: [1,1] +Explanation: A = 1, B = 1. A + B = n and both A and B don't contain any 0 in their decimal representation. +``` + +**Example 2**: + +``` +Input: n = 11 +Output: [2,9] +``` + +**Example 3**: + +``` +Input: n = 10000 +Output: [1,9999] +``` + +**Example 4**: + +``` +Input: n = 69 +Output: [1,68] +``` + +**Example 5**: + +``` +Input: n = 1010 +Output: [11,999] +``` + +**Constraints**: + +- `2 <= n <= 10^4` + +## 题目大意 + +「无零整数」是十进制表示中 不含任何 0 的正整数。给你一个整数 n，请你返回一个 由两个整数组成的列表 [A, B]，满足： + +- A 和 B 都是无零整数 +- A + B = n + +题目数据保证至少有一个有效的解决方案。如果存在多个有效解决方案，你可以返回其中任意一个。 + +## 解题思路 + +- 给定一个整数 n，要求把它分解为 2 个十进制位中不含 0 的正整数且这两个正整数之和为 n。 +- 简单题。在 [1, n/2] 区间内搜索，只要有一组满足条件的解就 break。题目保证了至少有一组解，并且多组解返回任意一组即可。 + +## 代码 + +```go + +package leetcode + +func getNoZeroIntegers(n int) []int { + noZeroPair := []int{} + for i := 1; i <= n/2; i++ { + if isNoZero(i) && isNoZero(n-i) { + noZeroPair = append(noZeroPair, []int{i, n - i}...) + break + } + } + return noZeroPair +} + +func isNoZero(n int) bool { + for n != 0 { + if n%10 == 0 { + return false + } + n /= 10 + } + return true +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence.md b/website/content/ChapterFour/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence.md new file mode 100644 index 00000000..c8081e82 --- /dev/null +++ b/website/content/ChapterFour/1455.Check-If-a-Word-Occurs-As-a-Prefix-of-Any-Word-in-a-Sentence.md @@ -0,0 +1,98 @@ +# [1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence](https://leetcode.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/) + + +## 题目 + +Given a `sentence` that consists of some words separated by a **single space**, and a `searchWord`. + +You have to check if `searchWord` is a prefix of any word in `sentence`. + +Return *the index of the word* in `sentence` where `searchWord` is a prefix of this word (**1-indexed**). + +If `searchWord` is a prefix of more than one word, return the index of the first word **(minimum index)**. If there is no such word return **-1**. + +A **prefix** of a string `S` is any leading contiguous substring of `S`. + +**Example 1**: + +``` +Input: sentence = "i love eating burger", searchWord = "burg" +Output: 4 +Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence. + +``` + +**Example 2**: + +``` +Input: sentence = "this problem is an easy problem", searchWord = "pro" +Output: 2 +Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index. + +``` + +**Example 3**: + +``` +Input: sentence = "i am tired", searchWord = "you" +Output: -1 +Explanation: "you" is not a prefix of any word in the sentence. + +``` + +**Example 4**: + +``` +Input: sentence = "i use triple pillow", searchWord = "pill" +Output: 4 + +``` + +**Example 5**: + +``` +Input: sentence = "hello from the other side", searchWord = "they" +Output: -1 + +``` + +**Constraints**: + +- `1 <= sentence.length <= 100` +- `1 <= searchWord.length <= 10` +- `sentence` consists of lowercase English letters and spaces. +- `searchWord` consists of lowercase English letters. + +## 题目大意 + +给你一个字符串 sentence 作为句子并指定检索词为 searchWord ，其中句子由若干用 单个空格 分隔的单词组成。请你检查检索词 searchWord 是否为句子 sentence 中任意单词的前缀。 + +- 如果 searchWord 是某一个单词的前缀，则返回句子 sentence 中该单词所对应的下标（下标从 1 开始）。 +- 如果 searchWord 是多个单词的前缀，则返回匹配的第一个单词的下标（最小下标）。 +- 如果 searchWord 不是任何单词的前缀，则返回 -1 。 + +字符串 S 的 「前缀」是 S 的任何前导连续子字符串。 + +## 解题思路 + +- 给出 2 个字符串，一个是匹配串，另外一个是句子。在句子里面查找带匹配串前缀的单词，并返回第一个匹配单词的下标。 +- 简单题。按照题意，扫描一遍句子，一次匹配即可。 + +## 代码 + +```go + +package leetcode + +import "strings" + +func isPrefixOfWord(sentence string, searchWord string) int { + for i, v := range strings.Split(sentence, " ") { + if strings.HasPrefix(v, searchWord) { + return i + 1 + } + } + return -1 +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/1464.Maximum-Product-of-Two-Elements-in-an-Array.md b/website/content/ChapterFour/1464.Maximum-Product-of-Two-Elements-in-an-Array.md new file mode 100644 index 00000000..04dbf134 --- /dev/null +++ b/website/content/ChapterFour/1464.Maximum-Product-of-Two-Elements-in-an-Array.md @@ -0,0 +1,66 @@ +# [1464. Maximum Product of Two Elements in an Array](https://leetcode.com/problems/maximum-product-of-two-elements-in-an-array/) + + +## 题目 + +Given the array of integers `nums`, you will choose two different indices `i` and `j` of that array. Return the maximum value of `(nums[i]-1)*(nums[j]-1)`. + +**Example 1**: + +``` +Input: nums = [3,4,5,2] +Output: 12 +Explanation: If you choose the indices i=1 and j=2 (indexed from 0), you will get the maximum value, that is, (nums[1]-1)*(nums[2]-1) = (4-1)*(5-1) = 3*4 = 12. + +``` + +**Example 2**: + +``` +Input: nums = [1,5,4,5] +Output: 16 +Explanation: Choosing the indices i=1 and j=3 (indexed from 0), you will get the maximum value of (5-1)*(5-1) = 16. + +``` + +**Example 3**: + +``` +Input: nums = [3,7] +Output: 12 + +``` + +**Constraints**: + +- `2 <= nums.length <= 500` +- `1 <= nums[i] <= 10^3` + +## 题目大意 + +给你一个整数数组 nums，请你选择数组的两个不同下标 i 和 j，使 (nums[i]-1)*(nums[j]-1) 取得最大值。请你计算并返回该式的最大值。 + +## 解题思路 + +- 简单题。循环一次，按照题意动态维护 2 个最大值，从而也使得 `(nums[i]-1)*(nums[j]-1)` 能取到最大值。 + +## 代码 + +```go + +package leetcode + +func maxProduct(nums []int) int { + max1, max2 := 0, 0 + for _, num := range nums { + if num >= max1 { + max2 = max1 + max1 = num + } else if num <= max1 && num >= max2 { + max2 = num + } + } + return (max1 - 1) * (max2 - 1) +} + +``` \ No newline at end of file diff --git a/website/content/ChapterFour/1470.Shuffle-the-Array.md b/website/content/ChapterFour/1470.Shuffle-the-Array.md new file mode 100644 index 00000000..961d200c --- /dev/null +++ b/website/content/ChapterFour/1470.Shuffle-the-Array.md @@ -0,0 +1,64 @@ +# [1470. Shuffle the Array](https://leetcode.com/problems/shuffle-the-array/) + +## 题目 + +Given the array `nums` consisting of `2n` elements in the form `[x1,x2,...,xn,y1,y2,...,yn]`. + +*Return the array in the form* `[x1,y1,x2,y2,...,xn,yn]`. + +**Example 1**: + +``` +Input: nums = [2,5,1,3,4,7], n = 3 +Output: [2,3,5,4,1,7] +Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7]. + +``` + +**Example 2**: + +``` +Input: nums = [1,2,3,4,4,3,2,1], n = 4 +Output: [1,4,2,3,3,2,4,1] + +``` + +**Example 3**: + +``` +Input: nums = [1,1,2,2], n = 2 +Output: [1,2,1,2] + +``` + +**Constraints**: + +- `1 <= n <= 500` +- `nums.length == 2n` +- `1 <= nums[i] <= 10^3` + +## 题目大意 + +给你一个数组 nums ，数组中有 2n 个元素，按 [x1,x2,...,xn,y1,y2,...,yn] 的格式排列。请你将数组按 [x1,y1,x2,y2,...,xn,yn] 格式重新排列，返回重排后的数组。 + +## 解题思路 + +- 给定一个 2n 的数组，把后 n 个元素插空放到前 n 个元素里面。输出最终完成的数组。 +- 简单题，按照题意插空即可。 + +## 代码 + +```go + +package leetcode + +func shuffle(nums []int, n int) []int { + result := make([]int, 0) + for i := 0; i < n; i++ { + result = append(result, nums[i]) + result = append(result, nums[n+i]) + } + return result +} + +``` \ No newline at end of file