mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-06 09:23:19 +08:00
Add solution 6、8、12
This commit is contained in:
YDZ
26
leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion.go
Normal file
26
leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion.go
Normal file
@ -0,0 +1,26 @@
|
||||
package leetcode
|
||||
|
||||
func convert(s string, numRows int) string {
|
||||
matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
|
||||
for i := 0; i != len(s); {
|
||||
if down != numRows {
|
||||
matrix[down] = append(matrix[down], byte(s[i]))
|
||||
down++
|
||||
i++
|
||||
} else if up > 0 {
|
||||
matrix[up] = append(matrix[up], byte(s[i]))
|
||||
up--
|
||||
i++
|
||||
} else {
|
||||
up = numRows - 2
|
||||
down = 0
|
||||
}
|
||||
}
|
||||
solution := make([]byte, 0, len(s))
|
||||
for _, row := range matrix {
|
||||
for _, item := range row {
|
||||
solution = append(solution, item)
|
||||
}
|
||||
}
|
||||
return string(solution)
|
||||
}
|
||||
53
leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion_test.go
Normal file
53
leetcode/0006.ZigZag-Conversion/6. ZigZag Conversion_test.go
Normal file
@ -0,0 +1,53 @@
|
||||
package leetcode
|
||||
|
||||
import (
|
||||
"fmt"
|
||||
"testing"
|
||||
)
|
||||
|
||||
type question6 struct {
|
||||
para6
|
||||
ans6
|
||||
}
|
||||
|
||||
// para 是参数
|
||||
// one 代表第一个参数
|
||||
type para6 struct {
|
||||
s string
|
||||
numRows int
|
||||
}
|
||||
|
||||
// ans 是答案
|
||||
// one 代表第一个答案
|
||||
type ans6 struct {
|
||||
one string
|
||||
}
|
||||
|
||||
func Test_Problem6(t *testing.T) {
|
||||
|
||||
qs := []question6{
|
||||
|
||||
{
|
||||
para6{"PAYPALISHIRING", 3},
|
||||
ans6{"PAHNAPLSIIGYIR"},
|
||||
},
|
||||
|
||||
{
|
||||
para6{"PAYPALISHIRING", 4},
|
||||
ans6{"PINALSIGYAHRPI"},
|
||||
},
|
||||
|
||||
{
|
||||
para6{"A", 1},
|
||||
ans6{"A"},
|
||||
},
|
||||
}
|
||||
|
||||
fmt.Printf("------------------------Leetcode Problem 6------------------------\n")
|
||||
|
||||
for _, q := range qs {
|
||||
_, p := q.ans6, q.para6
|
||||
fmt.Printf("【input】:%v 【output】:%v\n", p, convert(p.s, p.numRows))
|
||||
}
|
||||
fmt.Printf("\n\n\n")
|
||||
}
|
||||
107
leetcode/0006.ZigZag-Conversion/README.md
Normal file
107
leetcode/0006.ZigZag-Conversion/README.md
Normal file
@ -0,0 +1,107 @@
|
||||
# [6. ZigZag Conversion](https://leetcode.com/problems/zigzag-conversion/)
|
||||
|
||||
|
||||
## 题目
|
||||
|
||||
The string `"PAYPALISHIRING"` is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
|
||||
|
||||
```
|
||||
P A H N
|
||||
A P L S I I G
|
||||
Y I R
|
||||
```
|
||||
|
||||
And then read line by line: `"PAHNAPLSIIGYIR"`
|
||||
|
||||
Write the code that will take a string and make this conversion given a number of rows:
|
||||
|
||||
```
|
||||
string convert(string s, int numRows);
|
||||
```
|
||||
|
||||
**Example 1:**
|
||||
|
||||
```
|
||||
Input: s = "PAYPALISHIRING", numRows = 3
|
||||
Output: "PAHNAPLSIIGYIR"
|
||||
```
|
||||
|
||||
**Example 2:**
|
||||
|
||||
```
|
||||
Input: s = "PAYPALISHIRING", numRows = 4
|
||||
Output: "PINALSIGYAHRPI"
|
||||
Explanation:
|
||||
P I N
|
||||
A L S I G
|
||||
Y A H R
|
||||
P I
|
||||
```
|
||||
|
||||
**Example 3:**
|
||||
|
||||
```
|
||||
Input: s = "A", numRows = 1
|
||||
Output: "A"
|
||||
```
|
||||
|
||||
**Constraints:**
|
||||
|
||||
- `1 <= s.length <= 1000`
|
||||
- `s` consists of English letters (lower-case and upper-case), `','` and `'.'`.
|
||||
- `1 <= numRows <= 1000`
|
||||
|
||||
## 题目大意
|
||||
|
||||
将一个给定字符串 `s` 根据给定的行数 `numRows` ，以从上往下、从左到右进行 Z 字形排列。
|
||||
|
||||
比如输入字符串为 `"PAYPALISHIRING"` 行数为 3 时，排列如下：
|
||||
|
||||
```go
|
||||
P A H N
|
||||
A P L S I I G
|
||||
Y I R
|
||||
```
|
||||
|
||||
之后，你的输出需要从左往右逐行读取，产生出一个新的字符串，比如：`"PAHNAPLSIIGYIR"`。
|
||||
|
||||
请你实现这个将字符串进行指定行数变换的函数：
|
||||
|
||||
```go
|
||||
string convert(string s, int numRows);
|
||||
```
|
||||
|
||||
## 解题思路
|
||||
|
||||
- 这一题没有什么算法思想，考察的是对程序控制的能力。用 2 个变量保存方向，当垂直输出的行数达到了规定的目标行数以后，需要从下往上转折到第一行，循环中控制好方向ji
|
||||
|
||||
## 代码
|
||||
|
||||
```go
|
||||
package leetcode
|
||||
|
||||
func convert(s string, numRows int) string {
|
||||
matrix, down, up := make([][]byte, numRows, numRows), 0, numRows-2
|
||||
for i := 0; i != len(s); {
|
||||
if down != numRows {
|
||||
matrix[down] = append(matrix[down], byte(s[i]))
|
||||
down++
|
||||
i++
|
||||
} else if up > 0 {
|
||||
matrix[up] = append(matrix[up], byte(s[i]))
|
||||
up--
|
||||
i++
|
||||
} else {
|
||||
up = numRows - 2
|
||||
down = 0
|
||||
}
|
||||
}
|
||||
solution := make([]byte, 0, len(s))
|
||||
for _, row := range matrix {
|
||||
for _, item := range row {
|
||||
solution = append(solution, item)
|
||||
}
|
||||
}
|
||||
return string(solution)
|
||||
}
|
||||
```
|
||||
@ -0,0 +1,56 @@
|
||||
package leetcode
|
||||
|
||||
func myAtoi(s string) int {
|
||||
maxInt, signAllowed, whitespaceAllowed, sign, digits := int64(2<<30), true, true, 1, []int{}
|
||||
for _, c := range s {
|
||||
if c == ' ' && whitespaceAllowed {
|
||||
continue
|
||||
}
|
||||
if signAllowed {
|
||||
if c == '+' {
|
||||
signAllowed = false
|
||||
whitespaceAllowed = false
|
||||
continue
|
||||
} else if c == '-' {
|
||||
sign = -1
|
||||
signAllowed = false
|
||||
whitespaceAllowed = false
|
||||
continue
|
||||
}
|
||||
}
|
||||
if c < '0' || c > '9' {
|
||||
break
|
||||
}
|
||||
whitespaceAllowed, signAllowed = false, false
|
||||
digits = append(digits, int(c-48))
|
||||
}
|
||||
var num, place int64
|
||||
place, num = 1, 0
|
||||
lastLeading0Index := -1
|
||||
for i, d := range digits {
|
||||
if d == 0 {
|
||||
lastLeading0Index = i
|
||||
} else {
|
||||
break
|
||||
}
|
||||
}
|
||||
if lastLeading0Index > -1 {
|
||||
digits = digits[lastLeading0Index+1:]
|
||||
}
|
||||
var rtnMax int64
|
||||
if sign > 0 {
|
||||
rtnMax = maxInt - 1
|
||||
} else {
|
||||
rtnMax = maxInt
|
||||
}
|
||||
digitsCount := len(digits)
|
||||
for i := digitsCount - 1; i >= 0; i-- {
|
||||
num += int64(digits[i]) * place
|
||||
place *= 10
|
||||
if digitsCount-i > 10 || num > rtnMax {
|
||||
return int(int64(sign) * rtnMax)
|
||||
}
|
||||
}
|
||||
num *= int64(sign)
|
||||
return int(num)
|
||||
}
|
||||
192
leetcode/0008.String-to-Integer-atoi/README.md
Normal file
192
leetcode/0008.String-to-Integer-atoi/README.md
Normal file
@ -0,0 +1,192 @@
|
||||
# [8. String to Integer (atoi)](https://leetcode.com/problems/string-to-integer-atoi/)
|
||||
|
||||
|
||||
## 题目
|
||||
|
||||
Implement the `myAtoi(string s)` function, which converts a string to a 32-bit signed integer (similar to C/C++'s `atoi` function).
|
||||
|
||||
The algorithm for `myAtoi(string s)` is as follows:
|
||||
|
||||
1. Read in and ignore any leading whitespace.
|
||||
2. Check if the next character (if not already at the end of the string) is `'-'` or `'+'`. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
|
||||
3. Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
|
||||
4. Convert these digits into an integer (i.e. `"123" -> 123`, `"0032" -> 32`). If no digits were read, then the integer is `0`. Change the sign as necessary (from step 2).
|
||||
5. If the integer is out of the 32-bit signed integer range `[-231, 231 - 1]`, then clamp the integer so that it remains in the range. Specifically, integers less than `231` should be clamped to `231`, and integers greater than `231 - 1` should be clamped to `231 - 1`.
|
||||
6. Return the integer as the final result.
|
||||
|
||||
**Note:**
|
||||
|
||||
- Only the space character `' '` is considered a whitespace character.
|
||||
- **Do not ignore** any characters other than the leading whitespace or the rest of the string after the digits.
|
||||
|
||||
**Example 1:**
|
||||
|
||||
```
|
||||
Input: s = "42"
|
||||
Output: 42
|
||||
Explanation: The underlined characters are what is read in, the caret is the current reader position.
|
||||
Step 1: "42" (no characters read because there is no leading whitespace)
|
||||
^
|
||||
Step 2: "42" (no characters read because there is neither a '-' nor '+')
|
||||
^
|
||||
Step 3: "42" ("42" is read in)
|
||||
^
|
||||
The parsed integer is 42.
|
||||
Since 42 is in the range [-231, 231 - 1], the final result is 42.
|
||||
|
||||
```
|
||||
|
||||
**Example 2:**
|
||||
|
||||
```
|
||||
Input: s = " -42"
|
||||
Output: -42
|
||||
Explanation:
|
||||
Step 1: " -42" (leading whitespace is read and ignored)
|
||||
^
|
||||
Step 2: " -42" ('-' is read, so the result should be negative)
|
||||
^
|
||||
Step 3: " -42" ("42" is read in)
|
||||
^
|
||||
The parsed integer is -42.
|
||||
Since -42 is in the range [-231, 231 - 1], the final result is -42.
|
||||
|
||||
```
|
||||
|
||||
**Example 3:**
|
||||
|
||||
```
|
||||
Input: s = "4193 with words"
|
||||
Output: 4193
|
||||
Explanation:
|
||||
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
|
||||
^
|
||||
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
|
||||
^
|
||||
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
|
||||
^
|
||||
The parsed integer is 4193.
|
||||
Since 4193 is in the range [-231, 231 - 1], the final result is 4193.
|
||||
|
||||
```
|
||||
|
||||
**Example 4:**
|
||||
|
||||
```
|
||||
Input: s = "words and 987"
|
||||
Output: 0
|
||||
Explanation:
|
||||
Step 1: "words and 987" (no characters read because there is no leading whitespace)
|
||||
^
|
||||
Step 2: "words and 987" (no characters read because there is neither a '-' nor '+')
|
||||
^
|
||||
Step 3: "words and 987" (reading stops immediately because there is a non-digit 'w')
|
||||
^
|
||||
The parsed integer is 0 because no digits were read.
|
||||
Since 0 is in the range [-231, 231 - 1], the final result is 0.
|
||||
|
||||
```
|
||||
|
||||
**Example 5:**
|
||||
|
||||
```
|
||||
Input: s = "-91283472332"
|
||||
Output: -2147483648
|
||||
Explanation:
|
||||
Step 1: "-91283472332" (no characters read because there is no leading whitespace)
|
||||
^
|
||||
Step 2: "-91283472332" ('-' is read, so the result should be negative)
|
||||
^
|
||||
Step 3: "-91283472332" ("91283472332" is read in)
|
||||
^
|
||||
The parsed integer is -91283472332.
|
||||
Since -91283472332 is less than the lower bound of the range [-231, 231 - 1], the final result is clamped to -231 = -2147483648.
|
||||
```
|
||||
|
||||
**Constraints:**
|
||||
|
||||
- `0 <= s.length <= 200`
|
||||
- `s` consists of English letters (lower-case and upper-case), digits (`0-9`), `' '`, `'+'`
|
||||
|
||||
## 题目大意
|
||||
|
||||
请你来实现一个 myAtoi(string s) 函数，使其能将字符串转换成一个 32 位有符号整数（类似 C/C++ 中的 atoi 函数）。
|
||||
|
||||
函数 myAtoi(string s) 的算法如下：
|
||||
|
||||
- 读入字符串并丢弃无用的前导空格
|
||||
- 检查下一个字符（假设还未到字符末尾）为正还是负号，读取该字符（如果有）。 确定最终结果是负数还是正数。 如果两者都不存在，则假定结果为正。
|
||||
- 读入下一个字符，直到到达下一个非数字字符或到达输入的结尾。字符串的其余部分将被忽略。
|
||||
- 将前面步骤读入的这些数字转换为整数（即，"123" -> 123， "0032" -> 32）。如果没有读入数字，则整数为 0 。必要时更改符号（从步骤 2 开始）。
|
||||
- 如果整数数超过 32 位有符号整数范围 [−231, 231 − 1] ，需要截断这个整数，使其保持在这个范围内。具体来说，小于 −231 的整数应该被固定为 −231 ，大于 231 − 1 的整数应该被固定为 231 − 1 。
|
||||
- 返回整数作为最终结果。
|
||||
|
||||
注意：
|
||||
|
||||
- 本题中的空白字符只包括空格字符 ' ' 。
|
||||
- 除前导空格或数字后的其余字符串外，请勿忽略 任何其他字符。
|
||||
|
||||
## 解题思路
|
||||
|
||||
- 这题是简单题。题目要求实现类似 `C++` 中 `atoi` 函数的功能。这个函数功能是将字符串类型的数字转成 `int` 类型数字。先去除字符串中的前导空格，并判断记录数字的符号。数字需要去掉前导 `0` 。最后将数字转换成数字类型，判断是否超过 `int` 类型的上限 `[-2^31, 2^31 - 1]`，如果超过上限，需要输出边界，即 `-2^31`，或者 `2^31 - 1`。
|
||||
|
||||
## 代码
|
||||
|
||||
```go
|
||||
package leetcode
|
||||
|
||||
func myAtoi(s string) int {
|
||||
maxInt, signAllowed, whitespaceAllowed, sign, digits := int64(2<<30), true, true, 1, []int{}
|
||||
for _, c := range s {
|
||||
if c == ' ' && whitespaceAllowed {
|
||||
continue
|
||||
}
|
||||
if signAllowed {
|
||||
if c == '+' {
|
||||
signAllowed = false
|
||||
whitespaceAllowed = false
|
||||
continue
|
||||
} else if c == '-' {
|
||||
sign = -1
|
||||
signAllowed = false
|
||||
whitespaceAllowed = false
|
||||
continue
|
||||
}
|
||||
}
|
||||
if c < '0' || c > '9' {
|
||||
break
|
||||
}
|
||||
whitespaceAllowed, signAllowed = false, false
|
||||
digits = append(digits, int(c-48))
|
||||
}
|
||||
var num, place int64
|
||||
place, num = 1, 0
|
||||
lastLeading0Index := -1
|
||||
for i, d := range digits {
|
||||
if d == 0 {
|
||||
lastLeading0Index = i
|
||||
} else {
|
||||
break
|
||||
}
|
||||
}
|
||||
if lastLeading0Index > -1 {
|
||||
digits = digits[lastLeading0Index+1:]
|
||||
}
|
||||
var rtnMax int64
|
||||
if sign > 0 {
|
||||
rtnMax = maxInt - 1
|
||||
} else {
|
||||
rtnMax = maxInt
|
||||
}
|
||||
digitsCount := len(digits)
|
||||
for i := digitsCount - 1; i >= 0; i-- {
|
||||
num += int64(digits[i]) * place
|
||||
place *= 10
|
||||
if digitsCount-i > 10 || num > rtnMax {
|
||||
return int(int64(sign) * rtnMax)
|
||||
}
|
||||
}
|
||||
num *= int64(sign)
|
||||
return int(num)
|
||||
}
|
||||
```
|
||||
15
leetcode/0012.Integer-to-Roman/12. Integer to Roman.go
Normal file
15
leetcode/0012.Integer-to-Roman/12. Integer to Roman.go
Normal file
@ -0,0 +1,15 @@
|
||||
package leetcode
|
||||
|
||||
func intToRoman(num int) string {
|
||||
values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
|
||||
symbols := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
|
||||
res, i := "", 0
|
||||
for num != 0 {
|
||||
for values[i] > num {
|
||||
i++
|
||||
}
|
||||
num -= values[i]
|
||||
res += symbols[i]
|
||||
}
|
||||
return res
|
||||
}
|
||||
71
leetcode/0012.Integer-to-Roman/12. Integer to Roman_test.go
Normal file
71
leetcode/0012.Integer-to-Roman/12. Integer to Roman_test.go
Normal file
@ -0,0 +1,71 @@
|
||||
package leetcode
|
||||
|
||||
import (
|
||||
"fmt"
|
||||
"testing"
|
||||
)
|
||||
|
||||
type question12 struct {
|
||||
para12
|
||||
ans12
|
||||
}
|
||||
|
||||
// para 是参数
|
||||
// one 代表第一个参数
|
||||
type para12 struct {
|
||||
one int
|
||||
}
|
||||
|
||||
// ans 是答案
|
||||
// one 代表第一个答案
|
||||
type ans12 struct {
|
||||
one string
|
||||
}
|
||||
|
||||
func Test_Problem12(t *testing.T) {
|
||||
|
||||
qs := []question12{
|
||||
|
||||
{
|
||||
para12{3},
|
||||
ans12{"III"},
|
||||
},
|
||||
|
||||
{
|
||||
para12{4},
|
||||
ans12{"IV"},
|
||||
},
|
||||
|
||||
{
|
||||
para12{9},
|
||||
ans12{"IX"},
|
||||
},
|
||||
|
||||
{
|
||||
para12{58},
|
||||
ans12{"LVIII"},
|
||||
},
|
||||
|
||||
{
|
||||
para12{1994},
|
||||
ans12{"MCMXCIV"},
|
||||
},
|
||||
{
|
||||
para12{123},
|
||||
ans12{"CXXIII"},
|
||||
},
|
||||
|
||||
{
|
||||
para12{120},
|
||||
ans12{"CXX"},
|
||||
},
|
||||
}
|
||||
|
||||
fmt.Printf("------------------------Leetcode Problem 12------------------------\n")
|
||||
|
||||
for _, q := range qs {
|
||||
_, p := q.ans12, q.para12
|
||||
fmt.Printf("【input】:%v 【output】:%v\n", p.one, intToRoman(p.one))
|
||||
}
|
||||
fmt.Printf("\n\n\n")
|
||||
}
|
||||
102
leetcode/0012.Integer-to-Roman/README.md
Normal file
102
leetcode/0012.Integer-to-Roman/README.md
Normal file
@ -0,0 +1,102 @@
|
||||
# [12. Integer to Roman](https://leetcode.com/problems/integer-to-roman/)
|
||||
|
||||
|
||||
## 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, `2` is written as `II` in Roman numeral, just two one's added together. `12` is written as `XII`, which is simply `X + II`. The number `27` 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 an integer, convert it to a roman numeral.
|
||||
|
||||
**Example 1:**
|
||||
|
||||
```
|
||||
Input: num = 3
|
||||
Output: "III"
|
||||
```
|
||||
|
||||
**Example 2:**
|
||||
|
||||
```
|
||||
Input: num = 4
|
||||
Output: "IV"
|
||||
```
|
||||
|
||||
**Example 3:**
|
||||
|
||||
```
|
||||
Input: num = 9
|
||||
Output: "IX"
|
||||
```
|
||||
|
||||
**Example 4:**
|
||||
|
||||
```
|
||||
Input: num = 58
|
||||
Output: "LVIII"
|
||||
Explanation: L = 50, V = 5, III = 3.
|
||||
```
|
||||
|
||||
**Example 5:**
|
||||
|
||||
```
|
||||
Input: num = 1994
|
||||
Output: "MCMXCIV"
|
||||
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.
|
||||
```
|
||||
|
||||
**Constraints:**
|
||||
|
||||
- `1 <= num <= 3999`
|
||||
|
||||
## 题目大意
|
||||
|
||||
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。
|
||||
|
||||
给定一个整数，将其转为罗马数字。输入确保在 1 到 3999 的范围内。
|
||||
|
||||
## 解题思路
|
||||
|
||||
- 依照题意，优先选择大的数字，解题思路采用贪心算法。将 1-3999 范围内的罗马数字从大到小放在数组中，从头选择到尾，即可把整数转成罗马数字。
|
||||
|
||||
## 代码
|
||||
|
||||
```go
|
||||
package leetcode
|
||||
|
||||
func intToRoman(num int) string {
|
||||
values := []int{1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1}
|
||||
symbols := []string{"M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"}
|
||||
res, i := "", 0
|
||||
for num != 0 {
|
||||
for values[i] > num {
|
||||
i++
|
||||
}
|
||||
num -= values[i]
|
||||
res += symbols[i]
|
||||
}
|
||||
return res
|
||||
}
|
||||
```
|
||||
Reference in New Issue
Block a user