<p>Given an array of integers <code>nums</code> and an integer <code>k</code>, return <em>the number of <b>unique</b> k-diff pairs in the array</em>.</p>

<p>A <strong>k-diff</strong> pair is an integer pair <code>(nums[i], nums[j])</code>, where the following are true:</p>

<ul>
	<li><code>0 &lt;= i, j &lt; nums.length</code></li>
	<li><code>i != j</code></li>
	<li><code>|nums[i] - nums[j]| == k</code></li>
</ul>

<p><strong>Notice</strong> that <code>|val|</code> denotes the absolute value of <code>val</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,1,4,1,5], k = 2
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of <strong>unique</strong> pairs.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5], k = 1
<strong>Output:</strong> 4
<strong>Explanation:</strong> There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,3,1,5,4], k = 0
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is one 0-diff pair in the array, (1, 1).
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>7</sup> &lt;= nums[i] &lt;= 10<sup>7</sup></code></li>
	<li><code>0 &lt;= k &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution {
public:
    int findPairs(vector<int>& nums, int k);
};
/*******************************************************************/
int Solution::findPairs(vector<int>& nums, int k) {
    int count=0;
    if (k == 0) {
        int i, size=nums.size();
        map<int,int> m;
        map<int,int>::iterator itm;
        for (i = 0; i < size; ++i) {
            ++m[nums[i]];
        }
        for (itm = m.begin(); itm != m.end(); ++itm) {
            if (itm->second > 1) {
                ++count;
            }
        }
        return count;
    }
    set<int> s(nums.begin(), nums.end());
    set<int>::iterator it1, it2, ref1 = s.end(), ref2;
    --ref1;
    for (it1 = s.begin(); it1 != ref1; ++it1) {
        ref2 = it1;
        ++ref2;
        for(it2 = ref2; it2 != s.end(); ++it2) {
            if (*it2-*it1 == k) {
                ++count;
            } else if (*it2-*it1 > k ) {
                break;
            }
        }
    }
    return count;
}
/*******************************************************************/
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