922. Sort Array By Parity II （easy）

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.

 題目要求偶數在偶數位，奇數在奇數位，順序可以隨便，也不要求inplace

1 奇偶分離再錯位插入（not fast

class Solution:
    def sortArrayByParityII(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        '''search odd num in left & odd num in right'''
        odd = []
        even = []
        for x in A:
            if x % 2 == 0:even.append(x)
            else:odd.append(x)
        A.clear()
        for idx,x in enumerate(even):A.insert(idx*2, even[idx])
        for idx,x in enumerate(odd):A.insert(idx*2 + 1, odd[idx])
        return  A
        
 

untime: 236 ms, faster than 25.10% of Python3

2 two points判斷時就插入（fast

class Solution:
    def sortArrayByParityII(self, A):
        """
        :type A: List[int]
        :rtype: List[int]
        """
        even = 0
        odd = 1
        l = A.copy()
        length = len(A)
        for x in A:
            if x % 2 == 0:
                l[even] = x
                if(even < length - 2):even += 2
            else:
                l[odd] = x
                if(odd < length):odd += 2
        return  l
 

Runtime: 140 ms, faster than 95.16% of Python3