Leetcode：905. Sort array by parity [double pointer + implementation in three languages]

Title Description

Give you an array of integers nums, take nums Move all even elements in the array to the front of the array , Followed by all odd elements .

Returns... That meets this condition Any array As the answer .

Input and output

Input ：nums = [3,1,2,4]
Output ：[2,4,3,1]
explain ：[4,2,3,1]、[2,4,1,3] and [4,2,1,3] It will also be seen as the right answer .

Tips

• 1 <= nums.length <= 5000
• 0 <= nums[i] <= 5000

Ideas

Using double pointers to achieve , Similar to quick sort idea , Scan from the left until you encounter an odd number , Scan from the right until you encounter an even number , And then exchange the two numbers .

AC Code （C++）

class Solution {
    
public:
    vector<int> sortArrayByParity(vector<int>& nums) {
    
        int n = nums.size();
        int l = 0, r = n - 1;
        while(l < r) {
    
            while(l < r && nums[l] % 2 == 0) ++l;
            while(l < r && nums[r] % 2 == 1) --r;
            if (l < r) {
    
                swap(nums[l], nums[r]);
            }
        }
        return nums;
    }
};

AC Code （Java）

class Solution {
    
    public int[] sortArrayByParity(int[] nums) {
    
        int n = nums.length;
        int l = 0, r = n - 1;
        while(l < r) {
    
            while(l < r && nums[l] % 2 == 0) ++l;
            while(l < r && nums[r] % 2 == 1) --r;
            if (l < r) {
    
                nums[l] = nums[l] ^ nums[r];
                nums[r] = nums[l] ^ nums[r];
                nums[l] = nums[l] ^ nums[r];
            }
            
        }
        return nums;
    }
}

AC Code （Python）

class Solution:
    def sortArrayByParity(self, nums: List[int]) -> List[int]:
        n = len(nums)
        l, r = 0, n - 1
        while l < r:
            while l < r and nums[l] % 2 == 0:
                l+=1
            while l < r and nums[r] % 2 == 1:
                r-=1
            if l < r:
                nums[l], nums[r] = nums[r], nums[l]
        return nums