922. Sort Array By Parity II

題目敘述

Given an array of integers nums, half of the integers in nums are odd, and the other half are even.

Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.

Return any answer array that satisfies this condition.

Example 1:

Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.

Example 2:

Input: nums = [2,3]
Output: [2,3]

Constraints:

2 <= nums.length <= 2 * 10^4
nums.length is even.
Half of the integers in nums are even.
0 <= nums[i] <= 1000

Follow Up: Could you solve it in-place?

題目翻譯

這題是會可以個整數的陣列 nums，其中陣列的元素會有一半是偶數一半是奇數。然後要重新排列陣列內的元素，讓 index 是奇數的時候，其值也是奇數。反之亦然。

解法解析

此體可以使用一次的迴圈去遍歷，判斷奇偶數後，重新排列。因應不同程式語言的特性，有的程式語言會需要額外的陣列來記錄排列後的結果。有的則不需要（例如 Python 和 Go）。

程式範例

Python

class Solution:
    def sortArrayByParityII(self, nums: List[int]) -> List[int]:
        even_idx = 0
        odd_idx = 1
        sz = len(nums)

        while even_idx < sz and odd_idx < sz:
            if nums[even_idx] % 2 == 0:
                even_idx += 2
            elif nums[odd_idx] % 2 == 1:
                odd_idx += 2
            else:
                nums[even_idx], nums[odd_idx] = nums[odd_idx], nums[even_idx]
                even_idx += 2
                odd_idx += 2

        return nums

JavaScript

/**
 * @param {number[]} nums
 * @return {number[]}
 */
var sortArrayByParityII = function (nums) {
    let evenIdx = 0;
    let oddIdx = 1;

    while (evenIdx < nums.length && oddIdx < nums.length) {
        if (nums[evenIdx] % 2 === 0) {
            evenIdx += 2;
        } else if (nums[oddIdx] % 2 === 1) {
            oddIdx += 2;
        } else {
            let temp = nums[evenIdx];
            nums[evenIdx] = nums[oddIdx];
            nums[oddIdx] = temp;
            evenIdx += 2;
            oddIdx += 2;
        }
    }
    return nums;
};

Go

func sortArrayByParityII(nums []int) []int {
    length := len(nums)
    even_idx := 0
    odd_idx := 1

    for even_idx < length && odd_idx < length {
        if nums[even_idx]%2 == 0 {
            even_idx += 2
        } else if nums[odd_idx]%2 == 1 {
            odd_idx += 2
        } else {
            nums[even_idx], nums[odd_idx] = nums[odd_idx], nums[even_idx]
        }
    }
    return nums
}

Swift

class Solution {
    func sortArrayByParityII(_ nums: [Int]) -> [Int] {
        var result = [Int](repeating: 0, count: nums.count)
        var even = 0
        var odd = 1

        for num in nums{
            if num%2 == 0{
                result[even] = num
                even += 2
            }else{
                result[odd] = num
                odd += 2
            }
        }
        return result
    }
}

Kotlin

class Solution {
    fun sortArrayByParityII(nums: IntArray): IntArray {
        val result = IntArray(nums.size)
        var even = 0
        var odd = 1

        for (i in 0 until nums.size) {
            if (nums[i].rem(2) == 0) {
                result[even] = nums[i]
                even += 2
            } else {
                result[odd] = nums[i]
                odd += 2
            }
        }
        return result
    }
}

LeetCode Solution, Easy, 922. Sort Array By Parity II

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922. Sort Array By Parity II

題目敘述

題目翻譯

解法解析

程式範例

Python

JavaScript

Go

Swift

Kotlin

Command Palette

Comments

LeetCode Solution

More from this blog

922. Sort Array By Parity II

題目敘述

題目翻譯

解法解析

程式範例

Python

JavaScript

Go

Swift

Kotlin