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k-diff-pairs-in-an-array

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K-diff Pairs in an Array – LeetCode 532

Problem

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.

Answer

Original

Code

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class Solution {
public:
    int findPairs(vector<int>& nums, int k) {
        if(k < 0) return 0;
        sort(nums.begin(), nums.end());
        int cnt = 0;
        if(k){
            for(auto p = nums.cbegin(); p != nums.cend(); ++p){
                while((p + 1) != nums.cend() && *(p + 1) == *p) ++p;
                cnt += binary_search(p+1, nums.cend(), *p + k);
            }
        } else {
            bool flag = false;
            for(auto p = nums.cbegin(); p != nums.cend(); ++p){
                while((p + 1) != nums.cend() && *(p + 1) == *p){
                    ++p;
                    flag = true;
                }
                if(flag){
                    ++cnt;
                    flag = false;
                }
            }
        }
        return cnt;
    }
};

思路

要求特定差的数对，为了节省空间优先排序。然后保证单项次序下从前向后查找理想值。时间复杂度$O(nlog(n))$，空间复杂度$O(1)$。
耗时$16$ ms，排名$1.98\%$

Better

思路

通过存储对应值的出现次数，借助hash表做到对于期望值接近于常数时间的查询。典型的用空间换时间，具体耗时取决于数据结构的实现。时间复杂度$O(n)$，空间复杂度$O(n)$。
耗时$20$ ms，排名$6.87\%$

Code

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