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 <= 5000
0 <= A[i] <= 5000
分析:
将数组按照奇偶性分类,因为不要求顺序,用快排的partition函数即可。代码:
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
int l = 0, r = A.size()-1;
int temp = A[l];
while(l < r)
{
while((l<r)&&(A[r]%2)) --r;
A[l] = A[r];
while((l<r)&&!(A[l]%2)) ++l;
A[r] = A[l];
}
A[l] = temp;
return A;
}
};
上面代码跑了28ms,应该不是最快的,因为是新题目,还没看到别人的思路,有了新方法会更新。