leetcode 922. Sort Array By Parity II 按奇偶排序数组 II-耗时100

Problem: 922. Sort Array By Parity II 按奇偶排序数组 II

耗时100%，一个指针就行了，若满足题意则l++，r = l+1，若满足题意或者nums[l]和nums[r]奇偶性相同则r++，此时一定满足nums[l]和nums[r]不符合题意且奇偶性不同，然后交换两者且l+1，只需要遍历一次

Code

class Solution { public: vector<int> sortArrayByParityII(vector<int>& nums) { int n = nums.size(); int l = 0, r = 1; while(l < n) { while( l < n && (nums[l]&1)==(l&1) ) l++; r = l + 1; while( r < n && ( (nums[r]&1)==(r&1) || (nums[l]&1)==(nums[r]&1) ) ) r++; if(l < n && r < n) swap(nums[l++], nums[r]); } return nums; } };