344. 反转字符串
题目描述
编写一个函数,其作用是将输入的字符串反转过来。输入字符串以字符数组 char[] 的形式给出。
不要给另外的数组分配额外的空间,你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。
你可以假设数组中的所有字符都是 ASCII 码表中的可打印字符。
示例 1:
输入:["h","e","l","l","o"]输出:["o","l","l","e","h"]
示例 2:
输入:["H","a","n","n","a","h"]
输出:["h","a","n","n","a","H"]
解法
思路
- 双指针,left=0, right=len-1
- 循环条件为 left<right
- left和right调转
- left++ right—
实现
class Solution {
public void reverseString(char[] s) {
int left = 0;
int right = s.length - 1;
while (left < right) {
char tmp = s[left];
s[left] = s[right];
s[right] = tmp;
left++;
right--;
}
}
}
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
left, right = 0, len(s) - 1
while left < right:
s[left], s[right] = s[right], s[left]
left += 1
right -= 1
func reverseString(s []byte) {
left, right := 0, len(s)-1
for left < right {
s[left], s[right] = s[right], s[left]
left++
right--
}
return s
}
