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 <= A.length <= 50000 <= 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; } };
