README_EN.md

905. Sort Array By Parity

中文文档

Description

Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.

Return any array that satisfies this condition.

 

Example 1:

Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.

Example 2:

Input: nums = [0]
Output: [0]

 

Constraints:

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

Solutions

Python3

class Solution:
    def sortArrayByParity(self, nums: List[int]) -> List[int]:
        i, j = 0, len(nums) - 1
        while i < j:
            if nums[i] & 1:
                nums[i], nums[j] = nums[j], nums[i]
                j -= 1
            else:
                i += 1
        return nums

Java

class Solution {
    public int[] sortArrayByParity(int[] nums) {
        for (int i = 0, j = nums.length - 1; i < j;) {
            if (nums[i] % 2 == 1) {
                int t = nums[i];
                nums[i] = nums[j];
                nums[j] = t;
                --j;
            } else {
                ++i;
            }
        }
        return nums;
    }
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortArrayByParity = function (nums) {
    for (let i = 0, j = nums.length - 1; i < j; ) {
        if (nums[i] & 1) {
            [nums[i], nums[j]] = [nums[j], nums[i]];
            --j;
        } else {
            ++i;
        }
    }
    return nums;
};

Rust

impl Solution {
    pub fn sort_array_by_parity(mut nums: Vec<i32>) -> Vec<i32> {
        let (mut l, mut r) = (0, nums.len() - 1);
        while l < r {
            while l < r && nums[l] & 1 == 0 {
                l += 1;
            }
            while l < r && nums[r] & 1 == 1 {
                r -= 1;
            }
            nums.swap(l, r);
        }
        nums
    }
}

C++

class Solution {
public:
    vector<int> sortArrayByParity(vector<int>& nums) {
        for (int i = 0, j = nums.size() - 1; i < j;) {
            if (nums[i] & 1)
                swap(nums[i], nums[j--]);
            else
                ++i;
        }
        return nums;
    }
};

Go

func sortArrayByParity(nums []int) []int {
	for i, j := 0, len(nums)-1; i < j; {
		if nums[i]%2 == 1 {
			nums[i], nums[j] = nums[j], nums[i]
			j--
		} else {
			i++
		}
	}
	return nums
}

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