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:
- 2 <= A.length <= 20000
- A.length % 2 == 0
- 0 <= A[i] <= 1000
class Solution:
def sortArrayByParityII(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
odd = []
even = []
for i in A:
if i%2==1:
odd.append(i)
else:
even.append(i)
pos = 0
for i in range(len(odd)):
A[pos] = even[i]
A[pos+1] = odd[i]
pos +=2
return A
原文地址:https://www.cnblogs.com/bernieloveslife/p/9797409.html
时间: 2024-09-28 08:11:39