K-diff Pairs in an Array

题面：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 = 2Output: 2Explanation: 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 = 1Output: 4Explanation: 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 = 0Output: 1Explanation: 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]

题解：刚开始分k是否为0分开讨论，不为0的时候先排序，找出相同元素，除去，然后根据和找出数量即可，k为0的时候单独讨论即可。代码冗长，后看到一种不必分类讨论的方法，综合其长，同贴出来。

代码1：

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

代码2：

class Solution {public:    int findPairs(vector<int>& nums, int k) {          int res = 0, n = nums.size(), j = 0;        sort(nums.begin(), nums.end());        for (int i = 0; i < n; ++i) {            int j = max(j, i + 1);            while (j < n && (long)nums[j] - nums[i] < k) ++j;            if (j < n && (long)nums[j] - nums[i] == k) ++res;            while (i < n - 1 && nums[i] == nums[i + 1]) ++i;        }        return res;    }};