LeetCode.532 K-diff Pairs in an Array

题目：

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].

分析：

class Solution {    public int findPairs(int[] nums, int k) {     //给定数组，返回其数值对，两个数的绝对距离是k（即两数的差值的绝对值为k）        //要求：元素相同（顺序无关）的为一对，数组长度不超过10000，所有数的的范围为Integer范围        //解法1:暴力解法             //解法1:暴力解法        Arrays.sort(nums);        if(k<0) return 0;        int count=0;        for(int i=0;i<nums.length-1;i++){            //跳过相同的,该题需要考虑两种连续元素情况，相同元素只取一次            while(i<nums.length-2&&nums[i]==nums[i+2]){                i++;            }            for(int j=i+1;j<nums.length;j++){                while(j!=nums.length-1&&nums[j]==nums[j+1]){                    j++;                }                if(nums[j]-nums[i]==k){                    //表明在相同元素边界倒数2个，下次i直接从不同的元素开始                    if(k==0){i++;}                    count++;                    break;                }else if(nums[j]-nums[i]>k){                    break;                }else if(nums[i]==nums[j]){                    //为了下一次i从不同元素开始                    i++;                }            }        }        return count;    }}