922. Sort Array By Parity II (Easy)
https://leetcode.com/problems/sort-array-by-parity-ii/
Given an array A
of non-negative integers, half of the integers in A are odd, and half of the integers are even.
Sort the array so that whenever A[i]
is odd, i
is odd; and whenever A[i]
is even, i
is even.
You may return any answer array that satisfies this condition.
Example 1:
Input: [4,2,5,7] Output: [4,5,2,7] Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
Note:
2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000
Solutions
class Solution {
private void swap(int[] A, int i, int j) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
public int[] sortArrayByParityII(int[] A) {
int size = A.length;
// stands for the even index pointer
int evenPtr = 0;
// stands for the odd index pointer
int oddPtr = 1;
for (int i = 0; i < size; i++) {
// find odd value at even index
while (evenPtr < size) {
if (A[evenPtr] % 2 != 0) {
break;
}
evenPtr += 2;
}
// find even value at odd index
while (oddPtr < size) {
if (A[oddPtr] % 2 != 1) {
break;
}
oddPtr += 2;
}
if (evenPtr < size && oddPtr < size) {
swap(A, evenPtr, oddPtr);
}
}
return A;
}
}