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"]
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/reverse-string
解题思路:
指针交换法,无论是奇数还是偶数的个数,都可以前后置换
鄙人答案:
/**
* @param {character[]} s
* @return {void} Do not return anything, modify s in-place instead.
*/
// var reverseString = function(s) {
// s= s.reverse()
// };
var reverseString = function(s) {
let a=0,b=s.length-1;
while(a<b){
var ans = s[a];
s[a]=s[b];
s[b]=ans;
a++;
b--;
}
};