<p>In a warehouse, there is a row of barcodes, where the <code>i<sup>th</sup></code> barcode is <code>barcodes[i]</code>.</p>

<p>Rearrange the barcodes so that no two adjacent barcodes are equal. You may return any answer, and it is guaranteed an answer exists.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> barcodes = [1,1,1,2,2,2]
<strong>Output:</strong> [2,1,2,1,2,1]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> barcodes = [1,1,1,1,2,2,3,3]
<strong>Output:</strong> [1,3,1,3,1,2,1,2]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= barcodes.length &lt;= 10000</code></li>
	<li><code>1 &lt;= barcodes[i] &lt;= 10000</code></li>
</ul>

class Solution {
public:
    vector<int> rearrangeBarcodes(vector<int>& barcodes) {
        vector<int>* v;
        int i, j, k, n, t;
        bool bSwap;
        v = new vector<int>(barcodes.begin(), barcodes.end());
        n = barcodes.size();
        do {
            bSwap = false;
            for (i = 1; i < n; i++)
                if ((*v)[i] == (*v)[i - 1]) {
                    j = i;
                    do
                        j = (j + 1) % n;
                    while ((*v)[j] == (*v)[i - 1]);
                    t = (*v)[i];
                    (*v)[i] = (*v)[j];
                    (*v)[j] = t;
                    bSwap = true;
                }
        } while (bSwap);
        return *v;
    }
};
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