Add solution 1652、1653

2025-07-07 09:54:57 +08:00 · 2020-11-27 20:20:20 +08:00
parent 5770d110d0
commit 0c4b373319
6 changed files with 423 additions and 0 deletions
--- a/leetcode/1652.Defuse-the-Bomb/1652.
+++ b/leetcode/1652.Defuse-the-Bomb/1652.
@ -0,0 +1,56 @@
+package leetcode
+
+func decrypt(code []int, k int) []int {
+	if k == 0 {
+		for i := 0; i < len(code); i++ {
+			code[i] = 0
+		}
+		return code
+	}
+	count, sum, res := k, 0, make([]int, len(code))
+	if k > 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i + 1; j < len(code); j++ {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count--
+			}
+			if count > 0 {
+				for j := 0; j < len(code); j++ {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count--
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	if k < 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i - 1; j >= 0; j-- {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count++
+			}
+			if count < 0 {
+				for j := len(code) - 1; j >= 0; j-- {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count++
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	return res
+}
--- a/leetcode/1652.Defuse-the-Bomb/1652.
+++ b/leetcode/1652.Defuse-the-Bomb/1652.
@ -0,0 +1,53 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1652 struct {
+	para1652
+	ans1652
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1652 struct {
+	code []int
+	k    int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1652 struct {
+	one []int
+}
+
+func Test_Problem1652(t *testing.T) {
+
+	qs := []question1652{
+
+		{
+			para1652{[]int{5, 7, 1, 4}, 3},
+			ans1652{[]int{12, 10, 16, 13}},
+		},
+
+		{
+			para1652{[]int{1, 2, 3, 4}, 0},
+			ans1652{[]int{0, 0, 0, 0}},
+		},
+
+		{
+			para1652{[]int{2, 4, 9, 3}, -2},
+			ans1652{[]int{12, 5, 6, 13}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1652------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1652, q.para1652
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, decrypt(p.code, p.k))
+	}
+	fmt.Printf("\n\n\n")
+}
--- a/leetcode/1652.Defuse-the-Bomb/README.md
+++ b/leetcode/1652.Defuse-the-Bomb/README.md
@ -0,0 +1,125 @@
+# [1652. Defuse the Bomb](https://leetcode.com/problems/defuse-the-bomb/)
+
+
+## 题目
+
+You have a bomb to defuse, and your time is running out! Your informer will provide you with a **circular** array `code` of length of `n` and a key `k`.
+
+To decrypt the code, you must replace every number. All the numbers are replaced **simultaneously**.
+
+- If `k > 0`, replace the `ith` number with the sum of the **next** `k` numbers.
+- If `k < 0`, replace the `ith` number with the sum of the **previous** `k` numbers.
+- If `k == 0`, replace the `ith` number with `0`.
+
+As `code` is circular, the next element of `code[n-1]` is `code[0]`, and the previous element of `code[0]` is `code[n-1]`.
+
+Given the **circular** array `code` and an integer key `k`, return *the decrypted code to defuse the bomb*!
+
+**Example 1:**
+
+```
+Input: code = [5,7,1,4], k = 3
+Output: [12,10,16,13]
+Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.
+```
+
+**Example 2:**
+
+```
+Input: code = [1,2,3,4], k = 0
+Output: [0,0,0,0]
+Explanation: When k is zero, the numbers are replaced by 0. 
+```
+
+**Example 3:**
+
+```
+Input: code = [2,4,9,3], k = -2
+Output: [12,5,6,13]
+Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.
+```
+
+**Constraints:**
+
+- `n == code.length`
+- `1 <= n <= 100`
+- `1 <= code[i] <= 100`
+- `(n - 1) <= k <= n - 1`
+
+## 题目大意
+
+你有一个炸弹需要拆除，时间紧迫！你的情报员会给你一个长度为 n 的 循环 数组 code 以及一个密钥 k 。为了获得正确的密码，你需要替换掉每一个数字。所有数字会 同时 被替换。
+
+- 如果 k > 0 ，将第 i 个数字用 接下来 k 个数字之和替换。
+- 如果 k < 0 ，将第 i 个数字用 之前 k 个数字之和替换。
+- 如果 k == 0 ，将第 i 个数字用 0 替换。
+
+由于 code 是循环的， code[n-1] 下一个元素是 code[0] ，且 code[0] 前一个元素是 code[n-1] 。
+
+给你 循环 数组 code 和整数密钥 k ，请你返回解密后的结果来拆除炸弹！
+
+## 解题思路
+
+- 给出一个 code 数组，要求按照规则替换每个字母。
+- 简单题，按照题意描述循环即可。
+
+## 代码
+
+```go
+package leetcode
+
+func decrypt(code []int, k int) []int {
+	if k == 0 {
+		for i := 0; i < len(code); i++ {
+			code[i] = 0
+		}
+		return code
+	}
+	count, sum, res := k, 0, make([]int, len(code))
+	if k > 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i + 1; j < len(code); j++ {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count--
+			}
+			if count > 0 {
+				for j := 0; j < len(code); j++ {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count--
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	if k < 0 {
+		for i := 0; i < len(code); i++ {
+			for j := i - 1; j >= 0; j-- {
+				if count == 0 {
+					break
+				}
+				sum += code[j]
+				count++
+			}
+			if count < 0 {
+				for j := len(code) - 1; j >= 0; j-- {
+					if count == 0 {
+						break
+					}
+					sum += code[j]
+					count++
+				}
+			}
+			res[i] = sum
+			sum, count = 0, k
+		}
+	}
+	return res
+}
+```
--- a/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653.
+++ b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653.
@ -0,0 +1,42 @@
+package leetcode
+
+// 解法一 DP
+func minimumDeletions(s string) int {
+	prev, res, bCount := 0, 0, 0
+	for _, c := range s {
+		if c == 'a' {
+			res = min(prev+1, bCount)
+			prev = res
+		} else {
+			bCount++
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+// 解法二 模拟
+func minimumDeletions1(s string) int {
+	aCount, bCount, res := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount++
+		}
+	}
+	res = aCount
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount--
+		} else {
+			bCount++
+		}
+		res = min(res, aCount+bCount)
+	}
+	return res
+}
--- a/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653.
+++ b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/1653.
@ -0,0 +1,57 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question1649 struct {
+	para1649
+	ans1649
+}
+
+// para 是参数
+// one 代表第一个参数
+type para1649 struct {
+	s string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans1649 struct {
+	one int
+}
+
+func Test_Problem1649(t *testing.T) {
+
+	qs := []question1649{
+
+		{
+			para1649{"aababbab"},
+			ans1649{2},
+		},
+
+		{
+			para1649{"bbaaaaabb"},
+			ans1649{2},
+		},
+
+		{
+			para1649{"b"},
+			ans1649{0},
+		},
+
+		{
+			para1649{"ababaaaabbbbbaaababbbbbbaaabbaababbabbbbaabbbbaabbabbabaabbbababaa"},
+			ans1649{25},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 1649------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans1649, q.para1649
+		fmt.Printf("【input】:%v      【output】:%v      \n", p, minimumDeletions(p.s))
+	}
+	fmt.Printf("\n\n\n")
+}
--- a/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/README.md
+++ b/leetcode/1653.Minimum-Deletions-to-Make-String-Balanced/README.md
@ -0,0 +1,90 @@
+# [1653. Minimum Deletions to Make String Balanced](https://leetcode.com/problems/minimum-deletions-to-make-string-balanced/)
+
+
+## 题目
+
+You are given a string `s` consisting only of characters `'a'` and `'b'`.
+
+You can delete any number of characters in `s` to make `s` **balanced**. `s` is **balanced** if there is no pair of indices `(i,j)` such that `i < j` and `s[i] = 'b'` and `s[j]= 'a'`.
+
+Return *the **minimum** number of deletions needed to make* `s` ***balanced***.
+
+**Example 1:**
+
+```
+Input: s = "aababbab"
+Output: 2
+Explanation: You can either:
+Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
+Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
+```
+
+**Example 2:**
+
+```
+Input: s = "bbaaaaabb"
+Output: 2
+Explanation: The only solution is to delete the first two characters.
+```
+
+**Constraints:**
+
+- `1 <= s.length <= 105`
+- `s[i]` is `'a'` or `'b'`.
+
+## 题目大意
+
+给你一个字符串 s ，它仅包含字符 'a' 和 'b' 。你可以删除 s 中任意数目的字符，使得 s 平衡 。我们称 s 平衡的 当不存在下标对 (i,j) 满足 i < j 且 s[i] = 'b' 同时 s[j]= 'a' 。请你返回使 s 平衡 的 最少 删除次数。
+
+## 解题思路
+
+- 给定一个字符串，要求删除最少次数，使得字母 a 都排在字母 b 的前面。
+- 很容易想到的一个解题思路是 DP。定义 `dp[i]` 为字符串下标 [ 0, i ] 这个区间内使得字符串平衡的最少删除次数。当 `s[i] == 'a'` 的时候，有 2 种情况，一种是 `s[i]` 前面全是 `[aa……aa]` 的情况，这个时候只需要把其中的所有的字母 `b` 删除即可。还有一种情况是 `s[i]` 前面有字母 `a` 也有字母 `b`，即 `[aaa……abb……b]`，这种情况就需要考虑 `dp[i-1]` 了。当前字母是 `a`，那么肯定要删除字母 `a`，来维持前面有一段字母 `b` 的情况。当 `s[i] == 'b'` 的时候，不管是 `[aa……aa]` 这种情况，还是 `[aaa……abb……b]` 这种情况，当前字母 `b` 都可以直接附加在后面，也能保证整个字符串是平衡的。所以状态转移方程为 `dp[i+1] = min(dp[i] + 1, bCount), s[i] == 'a'`，`dp[i+1] = dp[i], s[i] == 'b'`。最终答案存在 `dp[n]` 中。由于前后项的递推关系中只用到一次前一项，所以我们还可以优化一下空间，用一个变量保存前一项的结果。优化以后的代码见解法一。
+- 这一题还有一个模拟的思路。题目要求找到最小删除字数，那么就是要找到一个“临界点”，在这个临界点的左边删除所有的字母 b，在这个临界点的右边删除所有的字母 a。在所有的“临界点”中找到删除最少的次数。代码实现见解法二。
+
+## 代码
+
+```go
+package leetcode
+
+// 解法一 DP
+func minimumDeletions(s string) int {
+	prev, res, bCount := 0, 0, 0
+	for _, c := range s {
+		if c == 'a' {
+			res = min(prev+1, bCount)
+			prev = res
+		} else {
+			bCount++
+		}
+	}
+	return res
+}
+
+func min(a, b int) int {
+	if a < b {
+		return a
+	}
+	return b
+}
+
+// 解法二 模拟
+func minimumDeletions1(s string) int {
+	aCount, bCount, res := 0, 0, 0
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount++
+		}
+	}
+	res = aCount
+	for i := 0; i < len(s); i++ {
+		if s[i] == 'a' {
+			aCount--
+		} else {
+			bCount++
+		}
+		res = min(res, aCount+bCount)
+	}
+	return res
+}
+```