Leetcode922 sort array by parity II

Given an array of nonnegative integers  nums,  nums The middle half integer is Odd number , Half the whole number is even numbers .

Sort the array , For convenience  nums[i] In an odd number of ,i  It's also Odd number ; When  nums[i]  For even when , i It's also even numbers .

 Input ：nums = [4,2,5,7]
 Output ：[4,5,2,7]
 explain ：[4,7,2,5],[2,5,4,7],[2,7,4,5]  It will also be accepted .

Ideas ： Double pointer

If the original array can be modified , Then the local algorithm can be used to solve .

Maintain pointers for the even and odd subscript parts of the array respectively i, j. And then , At each step , If nums[i] It's odd , And keep moving forward j Move two units at a time , Until I meet the next even number . here , You can directly nums[i] And nums[j] In exchange for . We continue this process , Finally, all integers can be put in the correct position .

Implementation code ：

 public int[] sortArrayByParityII(int[] nums) {

    int j = 1;
    for(int i = 0; i < nums.length; i+=2){
        if(nums[i] % 2 == 1){

            while(nums[j] % 2 == 1){
                j += 2;
            }
            int temp = nums[i];
            nums[i] = nums[j];
            nums[j] = temp;
        }
    }
    return nums;
}