<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>

class Solution {
public:
    vector<int> sortArrayByParityII(vector<int>& nums) {
        int e=0,o=0,i=0;
        vector<int>v;
        while(v.size()!=nums.size())
        {
            while(e<nums.size() && nums[e]&1) e++;
                printf("e:%d\n",e);
                if(e<nums.size())
                {
                    v.push_back(nums[e]);
                    e++;
                }
            
            while(o<nums.size() && (nums[o]&1)==0) o++;
                printf("o:%d\n",o);
                if(o<nums.size())
                {
                    v.push_back(nums[o]);            
                    o++;
                }
        }
        return v;
    }
};
25200