# 922. Sort Array By Parity II 按奇偶排序数组 II

@TOC

## # 题目描述

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:

1. 2 <= A.length <= 20000
2. A.length % 2 == 0
3. 0 <= A[i] <= 1000

## # 解题方法

### # 使用奇偶数组

``````class Solution(object):
def sortArrayByParityII(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
odd = [x for x in A if x % 2 == 1]
even = [x for x in A if x % 2 == 0]
res = []
iseven = True
while odd or even:
if iseven:
res.append(even.pop())
else:
res.append(odd.pop())
iseven = not iseven
return res
``````

### # 排序

``````class Solution:
def sortArrayByParityII(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
A.sort(key = lambda x : x % 2)
N = len(A)
res = []
for i in range(N // 2):
res.append(A[i])
res.append(A[N - 1 - i])
return res
``````

### # 奇偶数位置变量

``````class Solution:
def sortArrayByParityII(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
N = len(A)
res = [0] * N
even, odd = 0, 1
for a in A:
if a % 2 == 1:
res[odd] = a
odd += 2
else:
res[even] = a
even += 2
return res
``````

## # 日期

2018 年 10 月 14 日 —— 周赛做出来3个题，开心 2018 年 11 月 5 日 —— 打了羽毛球，有点累