[刷题防痴呆] 0532 - 数组中的k-diff数对 (K-d

题目地址

https://leetcode.com/problems/k-diff-pairs-in-an-array/

题目描述

532. K-diff Pairs in an Array

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).
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

思路

• hashmap.
• 如果k==0, 统计map中count大于1的数的个数.
• 如果k!=0, 统计map中key + k的个数.

关键点

代码

• 语言支持：Java

class Solution {
    public int findPairs(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int num: nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }

        int count = 0;
        if (k == 0) {
            for (int key: map.keySet()) {
                if (map.get(key) > 1) {
                    count++;
                }
            }
        } else {
            for (int key: map.keySet()) {
                if (map.containsKey(key + k)) {
                    count++;
                }
            }
        }

        return count;
    }
}