LeetCode解题报告（178）-- 532. K-diff Pairs in an Array

Problem

Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.

A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:

0 <= i, j < nums.length
i != j
a <= b
b - a == 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:

1
2
3

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:

1
2
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:

1 2	Input: nums = [1,2,4,4,3,3,0,9,2,3], k = 3 Output: 2

Example 5:

1 2	Input: nums = [-1,-2,-3], k = 1 Output: 2

Constraints:

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

Analysis

这道题目要在给定的数组中找到差为k的数对，其实有点像是two sum的变种。假设数组中存在两个数a、b，使得a - b = k。如果在遍历到a的时候，我们只需要在数组中找一下是否存在a - k的数，如果存在，则说明这个是一个答案。这样下去就能够找到所有符合要求的答案了。

Solution

数据结构方面原来使用set和map是差不多的，因为绝大多数情况下我们只需要检查有没有这个数，而不考虑它出现的次数。但是有一种edge case需要考虑，就是当k为0的时候，两个相同的数是满足答案的要求的，所以要选用map，这样就是记录数字出现的次数了。

Code

class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        unordered_map<int, int> m;
        for (auto num: nums) {
            m[num]++;
        }
        
        int result = 0;
        int size = nums.size();
        for (auto temp: m) {
            if (k == 0 && temp.second > 1) {
                result++;
            } else if (k != 0 && m.count(temp.first - k)) {
                result++;
            }
        }
        return result;
    }
};

Summary

想这类在数组中找一对的题目，可以优先考虑使用map，因为一旦确定一个元素之后，另一个元素也是可以确定的。这道题的分析到这里，谢谢！