922. Sort Array By Parity II

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本文介绍了解决LeetCode上编号为922的问题Sort Array By Parity II的两种方法。第一种方法是将数组分为奇数和偶数两个子数组，然后重新组合。第二种方法是在原数组中通过检查和交换元素来达到目标条件。这两种方法都能有效地解决题目要求，即确保数组中偶数位置上的元素为偶数，奇数位置上的元素为奇数。

922. Sort Array By Parity II

Easy

题目

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.

Note:

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

解法

1.

比较无赖的解法，将整个数组分成两个子数组，分别存储奇数和偶数，然后再将两个子数组合并起来：

class Solution:
    def sortArrayByParityII(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        a=[]
        b=[]
        for i in A:
            if i%2==0:
                a.append(i)
            else:
                b.append(i)
        length=len(A)
        A=[]        
        for i in range(0,int(length/2)):
            A.append(a[i])
            A.append(b[i])
            
        return A

2

检测第偶数个或者奇数个数的值，是否符合要求，如果不符合就进行交换：

class Solution(object):
    def sortArrayByParityII(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        j = 1
        for i in range(0, len(A), 2):
            if A[i] % 2 == 0:
                continue
            while A[j] % 2 == 1:
                j += 2
            A[i], A[j] = A[j], A[i]
        return A