<p>Given an array of integers <code>nums</code>, half of the integers in <code>nums</code> are <strong>odd</strong>, and the other half are <strong>even</strong>.</p>

<p>Sort the array so that whenever <code>nums[i]</code> is odd, <code>i</code> is <strong>odd</strong>, and whenever <code>nums[i]</code> is even, <code>i</code> is <strong>even</strong>.</p>

<p>Return <em>any answer array that satisfies this condition</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,2,5,7]
<strong>Output:</strong> [4,5,2,7]
<strong>Explanation:</strong> [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,3]
<strong>Output:</strong> [2,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>nums.length</code> is even.</li>
	<li>Half of the integers in <code>nums</code> are even.</li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow Up:</strong> Could you solve it in-place?</p>

#include <math.h>
class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& nums) {
        vector<int> odd;
        vector<int> even;
        vector<int> result;
        vector<int>::iterator iter;
        for(iter = nums.begin(); iter < nums.end(); iter++){
            if(((result.size() + *iter) % 2) == 0){
                result.push_back(*iter);
            }
            else{
                if(*iter % 2 == 0){
                    even.push_back(*iter);
                }
                else{
                    odd.push_back(*iter);
                }
            }
        }
        if(!odd.empty() || !even.empty()){
            if(result.size() % 2 == 0){
                while(result.size() < nums.size()){
                    result.push_back(even.back());
                    even.pop_back();
                    if(!odd.empty()){
                        result.push_back(odd.back());                    
                        odd.pop_back();
                    }
                }
            }
            else{
                while(result.size() < nums.size()){
                    result.push_back(odd.back());
                    odd.pop_back();
                    if(!even.empty()){
                        result.push_back(even.back());                    
                        even.pop_back();
                    }
                }
            }
        }
        return result;       
    }
};
25900