[sword finger offer] 21 Adjust array order so that odd numbers precede even numbers

Interview questions 21- Adjust the array order so that the odd Numbers precede the even Numbers

One 、 subject

Two 、 Ideas

• initialization ：I,J Double pointer , Point to the array nums Left and right ends

• Circular exchange ： When i >= j Out of the loop

  • The pointer i When an odd number is encountered, execute i = i + 1 skip , Until you find an even number
  • The pointer J If an even number is encountered, execute j = j - 1 skip , Until you find an odd number
  • In exchange for Nums[i] and nums[j] value

• Return value ： Return modified nums Array

3、 ... and 、 Code

class Solution {
    
public:
    vector<int> exchange(vector<int>& nums) {
    
        int i = 0,j = nums.size() - 1;
        while(i < j)
        {
    
            while(i < j && (nums[i] & 1) == 1) i++;//  Description encountered an odd number   direct ++  And higher computational efficiency 
            while(i < j && (nums[j] & 1) == 0) j--; //  Description encountered an even number   direct --

            swap(nums[i],nums[j]);
        }

        return nums;
    }
};