leetcode--532. K-diff Pairs in an Array

532. K-diff Pairs in an Array

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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.

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题目大意就是给定你一个数组和整数k，输出数组中任意两数差值为k的数目。有一对算一对但是相同的 [2,3][3,2]只统计一次。

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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.

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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).

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题解，根据理解容易程度：

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解法1：类似于冒泡排序，不过要去重。
先对数组进行排序，这样就只用考虑一边的情况，例如：[1,2,1]，k=1。如果不排序，[1,2]，[2,1]都要进行考虑，这个怎么进行排除将是一个很麻烦的事情。
排序了之后[1,1,2],那么我们只要考虑第一层是否重复了。还不理解，那就看代码。

class Solution {
    public int findPairs(int[] nums, int k) {
        int count = 0;
         Arrays.sort(nums);//可以替换为快速排序
             for(int i = 0;i < nums.length - 1 ;i++){

                    for(int j = i + 1;j < nums.length;j++){
                           if(nums[j] - nums[i] == k ){//不知道能不能用Math.abs()
                                count++;
                                break;
                            }                   
                    }
                 //上边代码已经统计了一次，所以相同的向后移位
                 while(i < nums.length-1 && nums[i] == nums[i+1]){
                        i++;
                }

             }
            return count;  
    }
}