题目描述

• 有点像快排的交换操作
  

思路 & 代码

• 类似快排中的找法，具体见代码注释

class Solution {public int[] exchange(int[] nums) {// 双指针啦～int left = 0, right = nums.length - 1;while(left < right){// 两次循环：分别找到需要交换的奇数、偶数while(left < right && nums[left] % 2 == 1){left++;}while(left < right && nums[right] % 2 == 0){right--;}// 每次交换一对奇偶数if(left < right){int temp = nums[left];nums[left] = nums[right];nums[right] = temp;}}return nums;}
}

二刷

• 就类似快排

class Solution {public int[] exchange(int[] nums) {int left = 0, right = nums.length - 1;while(left < right) {while(left < right && nums[right] % 2 == 0) {right--;}while(left < right && nums[left] % 2 == 1) {left++;}if(left < right) {int temp = nums[left];nums[left] = nums[right];nums[right] = temp;}}return nums;}
}