Formatted question description: https://leetcode.ca/all/905.html
905. Sort Array By Parity (Easy)
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
Solution 1. STL
// OJ: https://leetcode.com/problems/sort-array-by-parity/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
sort(begin(A), end(A), [](int a, int b) { return a % 2 < b % 2; });
return A;
}
};
Solution 2. Two Pointers
// OJ: https://leetcode.com/problems/sort-array-by-parity/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
for (int i = 0, j = A.size() - 1; i < j; ++i, --j) {
while (i < j && A[i] % 2 == 0) ++i;
while (i < j && A[j] % 2 != 0) --j;
if (i < j) swap(A[i], A[j]);
}
return A;
}
};
Java
class Solution {
public int[] sortArrayByParity(int[] A) {
if (A == null || A.length < 2)
return A;
int length = A.length;
for (int i = 1; i < length; i++) {
int num = A[i];
if (num % 2 != 0)
continue;
int insertIndex = 0;
for (int j = i - 1; j >= 0; j--) {
if (A[j] % 2 == 0) {
insertIndex = j + 1;
break;
} else
A[j + 1] = A[j];
}
A[insertIndex] = num;
}
return A;
}
}