LeetCode-K-diff Pairs in an Array

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Description：
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:

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

题意：给定一个一维数组A,和一个整数k，找出数组中任意两个数字满足其差的绝对值为k，返回所有可能的不重复的对数；

解法：最简单的办法就是遍历所有的可能，找到满足其绝对值的为k的两个元素；这里需要注意的是我们要排除重复的对数，因此对于第一重循环已经利用过的数字我们可以直接跳过；

class Solution {
    public int findPairs(int[] nums, int k) {
        Arrays.sort(nums);
        int result = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                int dif = Math.abs(nums[i] - nums[j]);
                if (dif == k) {result++; break;}
                else if (dif > k) break;
            }
            while (i < nums.length - 1 && nums[i] == nums[i + 1]) i++;
        }
        return result;
    }
}