922. Sort Array By Parity II
1. Description
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.
2. Example
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.
3. Note
- 2 <= A.length <= 20000
- A.length % 2 == 0
- 0 <= A[i] <= 1000
4. Solutions
My Accepted Solution
n = m_array.size()
Time complexity: O(n)
Space complexity: O(1)
class Solution
{
public:
// vector<int> sortArrayByParityII(vector<int>& A)
vector<int> sortArrayByParityII(vector<int> &m_array)
{
for(int even = 0, odd = 1; even < m_array.size() && odd < m_array.size(); )
{
while(even < m_array.size() && ((m_array[even] & 1) == 0)) even += 2;
while(odd < m_array.size() && ((m_array[odd] & 1) == 1)) odd += 2;
if(even < m_array.size() && odd < m_array.size())
{
swap(m_array[even], m_array[odd]);
even += 2;
odd += 2;
}
}
return m_array;
}
};