leetcode解题报告（13）：K-diff Pairs in an Array

描述

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].

分析

一开始我是想用两个for循环做的，后来发现漏了各种边界情况，提交数次都无法ac，深感无力。后来0看了讨论区（https://discuss.leetcode.com/topic/81702/o-n-concise-solution-c ）的解法恍然大悟：用一个unordered_map即可解决！

用unordered_map<int,int>，key 值保存数组元素，value保存该元素个数。

因此只需遍历这个map，判断这个map里是否存在键值加上k后的键值即可。

唯一需要考虑的是k=0的情况，因为k=0时只有两个相等的元素才能满足。并且根据题意，即使有n对匹配元素(val1,val2)是k匹配的，也只算做一对。因此k=0时，只需要判断这个元素的个数即可，有两个及以上便将个数加1.

代码如下：

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        if(k < 0)return 0;
        int k_diff = 0;
        unordered_map<int,int>um;
        for(int i = 0; i != nums.size(); ++i)
            ++um[nums[i]];
        if(k != 0){
            for(auto it = um.begin(); it != um.end(); ++it)
                if(um.find(it->first + k) != um.end())
                    ++k_diff;
        }else{
            for(auto it = um.begin(); it != um.end(); ++it)
                if(it->second > 1)
                    ++k_diff;
        }
        return k_diff;
    }
};

心得

写个心得。

unordered_map真心好用啊，很多操作的复杂度也低，在大多数情况下简直是上上之选，以后解题的时候可以优先考虑用unordered_map！