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"]
方法一: 使用数组的 reverse
方法。
/*** @param {character[]} s* @return {void} Do not return anything, modify s in-place instead.*/var reverseString = function(s) {s.reverse()}
方法二: 指针碰撞
/*** @param {character[]} s* @return {void} Do not return anything, modify s in-place instead.*/var reverseString = function(s) {let left = 0, right = s.length - 1while (left <= right) {swap(s, left, right)left++right--}}/* 交换位置nums 数组, a, b 为下标*/var swap = function(nums, a, b) {const tmp = nums[a]nums[a] = nums[b]nums[b] = tmp}
125、167、345