leetcode 532. K-diff Pairs in an Array 相差为k的元素对

客官°小女子只卖身不卖艺 2022-06-03 03:38 110阅读 0赞

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:  
The pairs (i, j) and (j, i) count as the same pair.  
The length of the array won’t exceed 10,000.  
All the integers in the given input belong to the range: \[-1e7, 1e7\].

本题题意很简单，直接使用map记录出现次数，然后分k==0与否遍历即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    
    using namespace std;
    
    
    class Solution
    {
    public:
        int findPairs(vector<int>& nums, int k) 
        {
            if (k < 0)
                return 0;
    
            map<int, int> mmp;
            for (int i : nums)
            {
                if (mmp.find(i) == mmp.end())
                    mmp[i] = 1;
                else
                    mmp[i] += 1;
            }
    
            int count = 0;
            if (k == 0)
            {
                for (auto i : mmp)
                {
                    if (i.second >= 2)
                        count++;
                }
            }
            else
            {
                for (auto i : mmp)
                {
                    if (mmp.find(i.first + k) != mmp.end())
                        count++;
                }
            }
            return count;
        }
    };