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"]
双指针从两端开始替换
/**
* @param {character[]} s
* @return {void} Do not return anything, modify s in-place instead.
*/
var reverseString = function(s) {
let len = s.length;
let l = 0, r = len - 1;
while(l < r) {
let tmp = s[l]
s[l] = s[r]
s[r] = tmp
l++
r--
}
};
时间复杂度
空间复杂度