LeetCode532. K-diff Pairs in an Array

最新推荐文章于 2020-04-02 11:34:06 发布

紫魔戒

最新推荐文章于 2020-04-02 11:34:06 发布

阅读量392

点赞数

CC 4.0 BY-SA版权

分类专栏： C-C++ 文章标签： leetcode

本文链接：https://blog.youkuaiyun.com/qq_26499321/article/details/72934202

C-C++ 专栏收录该内容

9 篇文章

订阅专栏

本文介绍了一种算法，用于解决寻找数组中特定差值k的唯一数对数量的问题，并给出了具体的实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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).

my program

思路：本题主要需要理解题目中给出的k-diff这个新概念，明白了后就不难解出答案。
利用一个map储存数组中的数字，以及出现的次数。当k=0的时候，统计map中value大于1的key的数目，即为所求k-diff的个数；当k>0的时候查找 $key' = key+k$ 在map中是否存在，若存在即k-diff对，count+1

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        int count = 0;
        if (k < 0) return count;
        map<int, int> mp;
        for (auto number : nums) {
            mp[number]++;
        }
        map<int, int>::iterator it;
        if (k == 0) {
            for (it = mp.begin(); it != mp.end(); it++) {
                if (it->second > 1)
                    count++;
            }
        }
        else {
            for (it = mp.begin(); it != mp.end(); it++) {
                if (mp.find(it->first+k) != mp.end())
                    count++;
            }
        }
        return count;
    }
};