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