leetcode922 Sort Array By Parity II

"""
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.
"""
"""
提供三种方法，传送门：https://blog.csdn.net/fuxuemingzhu/article/details/83045735
1.直接使用两个数组分别存放奇数和偶数，然后结果就是在这两个里面来回的选取就好了。
"""

class Solution1:
    def sortArrayByParityII(self, A):
        odd = [x for x in A if x % 2 == 1]  #奇数入栈
        even = [x for x in A if x % 2 == 0] #偶数入栈
        res = []
        iseven = True  #！！！判断奇偶数
        while odd or even:
            if iseven:
                res.append(even.pop())  #偶数写入结果
            else:
                res.append(odd.pop())   #奇数写入结果
            iseven = not iseven   #！！！下一个变为偶数
        return res

"""
先对A进行排序，使得偶数都在前面，奇数都在后面，
然后每次从前从后各取一个数，然后放到结果里就好了
"""
class Solution2:
    def sortArrayByParityII(self, A):
        A.sort(key=lambda x: x % 2)#!!!此方法可以将偶数防前面，奇数放后面 [0, 1]排序
        N = len(A)
        res = []
        for i in range(N // 2):
            #" / "就表示 浮点数除法，返回浮点结果;" // "表示整数除法。
            res.append(A[i])  #添加偶数
            res.append(A[N - 1 - i]) #添加奇数
        return res

"""
先把结果数组创建好，
然后使用奇偶数两个变量保存位置，
然后判断A中的每个数字是奇数还是偶数，
对应放到奇偶位置就行了。
"""
class Solution3:
    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  #！！！关键点，每次加2
            else:
                res[even] = a
                even += 2  #！！！
        return res