class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
int i = 0;int j = A.size() - 1;
while(i < j){
while(i < A.size() && A[i] % 2 == 0){
i ++;
}
while(j >= 0 && A[j] % 2 == 1){
j --;
}
if(i < j){
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
i++;j--;
}
}
return A;
}
};
905. Sort Array By Parity
Easy
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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] <= 500