Sword refers to Offer—— 21. Adjust the array order so that odd numbers are in front of even numbers

The topic is as follows

Enter an integer array and implement a function to adjust the order of the numbers in the array so that all odd numbers are in the first half of the array, and all even numbers are in the second half of the array.

Example:

Input: nums = [1,2,3,4]
Output: [1,3,2,4]
Note: [3,1,2,4] is also one of the correct answers.

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Own algorithm and code

(The algorithm class is correctly executed and submitted, and the memory consumption is high, mainly in the creation of a new array to store data)

• Create a new array result[], and two flags firstand last, pointing to result [] in the first 0position, the first nums.length-1position.
• Traverse the array nums from back to front.
  • If nums[i] is an even number, add it to the last position of result[] and move last one position forward, that is last--;
  • If nums[i] is an odd number, add it to the first position of result[] and move last one position back, that is first++;
• Then return result[].

public class Solution {
    
    

    public static int[] exchange(int[] nums) {
    
    
        int[] result = new int[nums.length];
        int first = 0;
        int last = nums.length-1;
        for (int i = nums.length-1; i >= 0; i--){
    
    
            if (nums[i]%2 == 0){
    
    
                result[last] = nums[i];
                last--;
            }else {
    
    
                result[first] = nums[i];
                first++;
            }
        }
        return result;
    }

    public static void main(String[] args) {
    
    
        int[] nums = {
    
    1,2,3,4};
        int[] result = exchange(nums);
        for (int num : result){
    
    
            System.out.println(num);
        }
    }
}

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Reference algorithm and code

Double pointer

• Define head pointer leftand tail pointer right.
• left has been moved to the right until the value it points to is even
• right keeps moving to the left until the value it points to is an odd number
• Exchange nums[left]and nums[right].
• Repeat the above operations until left==right.

public class Solution {
    
    

    public static int[] exchange1(int[] nums) {
    
    

        int left = 0;
        int right = nums.length-1;
        int temp;
        while (left < right){
    
    
            if (nums[left]%2 != 0){
    
    	//奇数
                left++;
                continue;
            }
            if (nums[right]%2 == 0){
    
     //偶数
                right--;
                continue;
            }
            temp = nums[left];
            nums[left] = nums[right];
            nums[right] = temp;
        }
        return nums;
    }

    public static void main(String[] args) {
    
    
        int[] nums = {
    
    1,2,3,4};
        int[] result = exchange1(nums);
        for (int num : result){
    
    
            System.out.println(num);
        }
    }
}