LeetCode 905. Sort Array By Parity

905. Sort Array By Parity

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.

Example 1:

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.

Note:

1. 1 <= A.length <= 5000

2. 0 <= A[i] <= 5000

题目描述：题意是要求给数组排序，排序的原则是偶数在前，奇数在后。

题目分析：很简单，我们直接给数组分分类就好了，然后用一个新的数组去存取值就行了。

python

代码：

class Solution(object):
def sortArrayByParity(self, A):
"""
:type A: List[int]
:rtype: List[int]
"""
odd_list = []
even_list = []
final_list = []
A_length = len(A)
for i in range(A_length):
if A[i] % 2 == 0:
even_list.append(A[i])
else:
odd_list.append(A[i])

final_list = even_list + odd_list
return final_list

C++

代码：

class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
vector<int> final_list(A.size());
int count_odd = 0;
int count_even = 0;
for(int i = 0; i < A.size(); i++){
if(A[i] % 2 == 0){
count_even++;
}
else{
count_odd++;
}
}
vector<int> odd_list(count_odd);
vector<int> even_list(count_even);
int odd = 0;
int even = 0;
for(int i = 0; i < A.size(); i++){
if(A[i] % 2 == 0){
even_list[even++] = A[i];
}
else{
odd_list[odd++] = A[i];
}
}
final_list = even_list;
final_list.insert(final_list.end(),odd_list.begin(),odd_list.end());
return final_list;
}
};