按奇偶排序数组

class Solution:
    def sortArrayByParityII(self, A: List[int]) -> List[int]:
        n=int(len(A)/2)
        odd_index=[0]*n
        even_index=[0]*n
        count_odd=0
        count_even=0
        for i in range(0,len(A)):
            a = A[i]%2
            b = i%2
            if a == 0 and b != 0:
                
                odd_index[count_odd] = i
                count_odd = count_odd+1
            elif a != 0 and b == 0:
                even_index[count_even] = i
                count_even = count_even+1
        
        for i in range(0,count_even):
            c = A[odd_index[i]]
            A[odd_index[i]] = A[even_index[i]]
            A[even_index[i]]=c
        return A

小周版

小周老师的思路是，以下标为0,1两个位置出发，步长为2，遍历列表，比如找到下标为奇值为偶时就让另一个位置出发找标偶值奇，再交换两个下标的位置就大功告成啦！速度比我的方法要快一点

class Solution:
    def sortArrayByParityII(self, A: List[int]) -> List[int]:
        jj=0
        n = int(len(A)/2)
        for i in range(0,n):
            if A[2*i+1]%2 == 0:
                
                for j in range(jj,n):
                    if A[2*j]%2 == 1:
                        jj=j
                        a=A[2*i+1]
                        A[2*i+1] = A[2*j]
                        A[2*j] = a 
                        break
                    
        return A