[LeetCode] 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 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: 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

Constraints:

• 1 <= nums.length <= 104
• -107 <= nums[i] <= 107
• 0 <= k <= 107

数组中的K-diff数对。题意是给一个整数数组和一个整数K，请你求出数组中有多少对数字满足两个数字之间的差值是K。

思路是用hashmap记录每个数字和他们出现的次数。再次遍历数组，也要分两种情况讨论

• 如果K = 0，那么只要找到任何一个出现超过两次的数字，就res++
• 如果K > 0，那么当遍历到num[i]的时候，就去看hashmap中是否有nums[i] + k，有则res++

对于第二种情况，我第一次做的时候有试图去找nums[i] - k，后来发觉没必要。因为hashmap会遍历所有的key，所以结果不会丢失

时间O(n)

空间O(n)

Java实现

class Solution {
    public int findPairs(int[] nums, int k) {
        // corner case
        if (nums == null || nums.length == 0) {
            return 0;
        }

        // normal case
        HashMap<Integer, Integer> map = new HashMap<>();
        int res = 0;
        for (int num : nums) {
            map.put(num, map.getOrDefault(num, 0) + 1);
        }

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

LeetCode 题目总结