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).
- //这也是今日头条的一道编程题
- public int findPairs1(int[] nums, int k) {
- // k小于0无意义
- if (nums == null || nums.length == 0 || k < 0)
- return 0;
- //这里使用了HashMap可以达到去重的目的
- Map<Integer, Integer> map = new HashMap<>();
- int i = 0;
- for (int num : nums)
- map.put(num, i++);
- int res = 0;
- for (i = 0; i < nums.length; i++)
- // map.get(nums[i] + k) != i主要是防止k等于0的情况
- if (map.containsKey(nums[i] + k) && map.get(nums[i] + k) != i) {
- map.remove(nums[i] + k);
- res++;
- }
- return res;
- }