今天开始学Java 求数组里差值为k的数组对 去重

532. K-diff Pairs in an Array

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here ak-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute differenceis 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).

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1. //这也是今日头条的一道编程题
2. public int findPairs1(int[] nums, int k) {  
3.     // k小于0无意义  
4.     if (nums == null || nums.length == 0 || k < 0)  
5.         return 0;  
6.    //这里使用了HashMap可以达到去重的目的
7.     Map<Integer, Integer> map = new HashMap<>();  
8.     int i = 0;  
9.     for (int num : nums)  
10.         map.put(num, i++);  
11.     int res = 0;  
12.     for (i = 0; i < nums.length; i++)  
13.         // map.get(nums[i] + k) != i主要是防止k等于0的情况
14.         if (map.containsKey(nums[i] + k) && map.get(nums[i] + k) != i) {  
15.             map.remove(nums[i] + k);  
16.             res++;  
17.         }  
18.     return res;  
19. }