Adjust the array order so that odd numbers precede even numbers - two questions per day

The finger of the sword Offer 21. Adjust the array order so that the odd Numbers precede the even Numbers

It's time to get up and brush questions every morning , It's really interesting to brush questions , Two questions a day , Come on ️

Enter an array of integers , Implement a function to adjust the order of the Numbers in the array , Make all odd numbers in the first half of the array , All even numbers are in the second half of the array .

Example ：

 Input ：nums = [1,2,3,4]
 Output ：[1,3,2,4] 
 notes ：[3,1,2,4]  And one of the right answers .

Tips ：

1. 0 <= nums.length <= 50000
2. 0 <= nums[i] <= 10000

Ideas

Algorithm ： Double pointer .

First, because ordered arrays , So the basic judgment is double pointer , Judge whether the number is odd nums[left]%2==1, If the left pointer is an odd number, move it right , If the right pointer is even, move left , If two pointers , On the left is an even number , On the right is the odd number , Exchange with each other , This problem is over .

public class  Adjust the array order so that the odd Numbers precede the even Numbers  {
    
    public int[] exchange(int[] nums) {
    
              int left=0;
              int right=nums.length-1;
              int temp=0;
              while (left<right){
    
                  if (nums[left]%2==1){
    
                      left++;
                  }else if (nums[right]%2==0){
    
                      right--;
                  }else {
    
                      temp=nums[left];
                      nums[left]=nums[right];
                      nums[right]=temp;
                      left++;
                      right--;
                  }
              }
              return nums;
    }
}