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
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& A) {
int iE = 0;// only check the Odd position
for(int iO =1;iO<A.size();iO+=2)
{
if (A[iO] %2 == 0)
{
while(A[iE] %2 == 0)
{
iE +=2;
}
int tmp = A[iE];
A[iE] = A[iO];
A[iO] = tmp;
}
}
return A;
}
};
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