Leetcode——532. K-diff Pairs in an Array

题目原址

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

题目描述

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.

Example1:

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.

Example2:

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).

Example3:

Input: [1, 3, 1, 5, 4], k = 0

Output: 1

Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

1. The pairs (i, j) and (j, i) count as the same pair.

2. The length of the array won’t exceed 10,000.

3. All the integers in the given input belong to the range: [-1e7, 1e7].

解题思路

将数组中的元素作为键放在HashMap中，键对应的值为同一个元素出现的个数。因此放入集合中之前需要判断集合中是否有该键

k有两种可能的情况，k=0时和k>0时

当k = 0时，计算值大于等于2的个数

当k>0时，计算

键等于当前键 + k

的个数

AC代码

class Solution {
public int findPairs(int[] nums, int k) {
int count = 0;
if(nums.length == 0 || k < 0)
return count;
Map<Integer,Integer> map = new HashMap<Integer,Integer>();

for(int i:nums) {
map.put(i, map.getOrDefault(i, 0) + 1);
}

for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
if(k == 0) {
if(entry.getValue() >= 2) {
count ++;
}
}else {
if(map.containsKey(entry.getKey()+ k))
count ++;
}
}
return count;
}
}

感谢

https://discuss.leetcode.com/topic/81714/java-o-n-solution-one-hashmap-easy-to-understand