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"]
class Solution:
def reverseString(self, s: List[str]) -> None:
"""
Do not return anything, modify s in-place instead.
"""
s[:] = s[::-1]
class Solution {
public void reverseString(char[] s) {
int n;
if (s == null || (n = s.length) < 2) return;
int i = 0, j = n - 1;
while (i < j) {
char t = s[i];
s[i] = s[j];
s[j] = t;
++i;
--j;
}
}
}