Given an array A
of non-negative integers, return an array consisting of all the even elements of A
, followed by all the odd elements of A
.
You may return any answer array that satisfies this condition.
Input: [3,1,2,4]
Output: [2,4,3,1]
The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
1 <= A.length <= 5000
0 <= A[i] <= 5000
Nothing need to analyse.
[3,1,2,4]
class Solution(object):
def sortArrayByParity(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
i=0
for j in range(len(A)):
if A[j] % 2 == 0:
A[i], A[j] = A[j], A[i]
i += 1
return A
Noticing the outside ‘i=0‘, recall the Bubble Sorting。
Bubble Sorting uses two iterations to get a former and a later.
原文:https://www.cnblogs.com/sxuer/p/10658006.html