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
public class Solution {
public int[] SortArrayByParityII(int[] A) {
if(A.Length<=1){
return A;
}
int evenIndexWithOddValue = -1;
int oddIndexWithEvenValue= -1;
evenIndexWithOddValue= FindNextEvenIndexWithOddValue(A,0);
oddIndexWithEvenValue= FindNextOddIndexWithEvenValue(A,0);
while(evenIndexWithOddValue != -1 && oddIndexWithEvenValue != -1){
int temp = A[evenIndexWithOddValue];
A[evenIndexWithOddValue] = A[oddIndexWithEvenValue];
A[oddIndexWithEvenValue]=temp;
evenIndexWithOddValue= FindNextEvenIndexWithOddValue(A,evenIndexWithOddValue+1);
oddIndexWithEvenValue= FindNextOddIndexWithEvenValue(A,oddIndexWithEvenValue+1);
}
return A;
}
private int FindNextEvenIndexWithOddValue(int[] A, int start){
for(int i=start;i<A.Length;i++){
if(A[i]%2==1 && i%2==0){
return i;
}
}
return -1;
}
private int FindNextOddIndexWithEvenValue(int[] A, int start){
for(int i=start;i<A.Length;i++){
if(A[i]%2==0 && i%2==1){
return i;
}
}
return -1;
}
}Time Complexity: O(n^2)
Space Complexity: O(1)
