（leetcode题解）K-diff Pairs in an Array

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

Input: [3, 1, 4, 1, 5], k = 2
Output: 2
Explanation: 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 unique pairs.

Example 2:

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

Example 3:

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

Note:

1. The pairs (i, j) and (j, i) count as the same pair.
2. The length of the array won't exceed 10,000.
3. All the integers in the given input belong to the range: [-1e7, 1e7].

这道题题意是在数组里寻找差值固定整数对，计算其个数，多次重复只记一次。

直接的相法就是先排序，然后用两个指针指示两个数，通过计算他们差值，进行移动，C++实现如下：

 int findPairs(vector<int>& nums, int k) {
        sort(nums.begin(),nums.end());
        int count=0;
        int length=nums.size();
        for(int i=0,j=1;j<length;)
        {
            if(nums[j]-nums[i]==k)
            {    
                count++;
                i++;
                j++;
                while(i<length&&nums[i]==nums[i-1])
                {
                    i++;
                }   
                while(j<length&&nums[j]==nums[j-1])
                {
                    j++;
                }
                if(i==j)
                    j++;
            }
            else if(nums[j]-nums[i]<k)
            {
                j++;
            }
            else 
            {
                i++;
                if(i==j)
                    j++;
            }
        }
        return count;
    }

另一种解法，通过unordered_map统计，在通过计算map值中是否存在差值为k的量。特例是当k=0时，map存的值大于等于2就符合。

int findPairs(vector<int>& nums, int k) {
        int count=0;
        unordered_map<int,int> h_map;
        for(auto &i:nums)
            h_map[i]++;
        for(auto &a:h_map)
        {
            if(k==0&&a.second>1)
                count++;
            if(k>0&&h_map.count(a.first+k)>0)
            {
                count++;
            }
        }
        return count;
    }