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) {
for (int i = 0, j = s.length - 1; i < j; ++i, --j) {
char t = s[i];
s[i] = s[j];
s[j] = t;
}
}
}
class Solution {
public:
void reverseString(vector<char>& s) {
for (int i = 0, j = s.size() - 1; i < j; ++i, --j)
swap(s[i], s[j]);
}
};
func reverseString(s []byte) {
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
}
/**
* @param {character[]} s
* @return {void} Do not return anything, modify s in-place instead.
*/
var reverseString = function(s) {
for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
[s[i], s[j]] = [s[j], s[i]];
}
};