922. Sort Array By Parity II

Description

Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even.

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

You may return any answer array that satisfies this condition.

Example 1:

Input: [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.

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Note

2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000

Links

(en)https://leetcode.com/problems/sort-array-by-parity-ii
(中文)https://leetcode-cn.com/problems/sort-array-by-parity-ii

Solutions

Solution1

将输入的原始数组分割成两个数组，一个存放偶数 even，一个存放奇数 odd。再通过一次循环，重新合成 result 数组。

Java Code

我解题时的线上提交代码:

class Solution {
    /**
    time complexity: O(n)
    space complexity: O(n)
    */
    public int[] sortArrayByParityII(int[] A) {
        
        List<Integer> even = new ArrayList<>();
        List<Integer> odd = new ArrayList<>();
        
        for (int n : A) {
            if (n % 2 == 0)
                even.add(n);
            else 
                odd.add(n);
        }
        int[] result = new int[A.length];
        for (int i = 0; i < A.length; i++) {
            if (i % 2 == 0) {
                result[i] = even.remove(0);
            } else {
                result[i] = odd.remove(0);
            }
        }
        return result;
    }
}

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Submission status

submission1-status

可以看到，这一解决方法在速度方面有点糟糕。

# 922. Sort Array By Parity II

# Description

# Note

# Links

# Solutions

# Solution1