https://leetcode.com/problems/k-diff-pairs-in-an-array/#/description
This is a very simple question. With two-pointer method, the time complexity is O(nlogn), because you need to sort it first. With a hashmap, you could save the time to rearrange it and hence it is only O(n).
The key points of the question are the boundary conditions. I forgot to consider the cases where k = 0 and k < 0!
Codes with a hashmap:
class Solution { public: int findPairs(vector<int>& nums, int k) { if (nums.empty() || k < 0) return 0; int result = 0; unordered_map<int, int> set; for (int i: nums) { set[i]++; } if (k == 0) { for (auto i: set) { if (i.second > 1) result++; } } else { for (auto i: set) { if (set.count(i.first + k)) result++; } } return result; } };
原文：http://www.cnblogs.com/tekkaman-blade/p/6922404.html