532 K-diff Pairs in an Array

Leetcode

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

0 <= i < j < nums.length
|nums[i] - nums[j]| == k

Notice that |val| denotes the absolute value of val.

Example 1:

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

可以排好序用双指针

class Solution {
    public int findPairs(int[] nums, int k) {

        Arrays.sort(nums);
        int diff, total=0;
        int len=nums.length;
        int left=0,right=1;
        while(right<len){
            diff=nums[right]-nums[left];
            if(diff<k){
                right=next(nums, right);
            } else if (diff>k) {
                left=next(nums,left);
                if(left==right) right++; //left 不能和 right 相同
            } else {
                total++;
                left=next(nums,left);
                right=left+1;
            }
        }
        return total;
    }

    // 获取下一个数
    private int next(int[] nums, int i){
        while(i+1<nums.length && nums[i+1]==nums[i]) ++i;
        return i+1;
    }
    
    
}