【leetcode】532. K-diff Pairs in an Array

532. K-diff Pairs in an Array

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

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].

思路

遍历数组，用HashMap储存数组，Key是数字，value是每个数字的数量，然后此时在hashmap中的entrySet中，数字已经是从小到大排列好了的，再循环这个entry数组，找到entry.getKey()+target的值。

代码

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

int count = 0;
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;
}
}