LeetCode 532 K-diff Pairs in an Array

题目

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
i != j
a <= b
b - a == k

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).

Example 4:
Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3
Output: 2

Example 5:
Input: nums = [-1,-2,-3], k = 1
Output: 2

Constraints:
1 <= nums.length <= 104
-107 <= nums[i] <= 107
0 <= k <= 107

思路

先将数组排序，对每一个nums[i]去右侧寻找是否有数可以构成k-pair，如果出现差值已经大于k的情况直接break（因为后面的数更大）。while (i >= 1 && i < nums.length && nums[i] == nums[i - 1]) i++;是为了处理重复元素的时候只有第一次计算，后面跳过，防止多算。

代码

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