Two Pointers/hash/3Sum/4Sum类题目

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当结果不唯一的时候，使用STL去重的最简单方式是

1 std::sort(res.begin(), res.end());2 res.erase(unique(res.begin(), res.end()), res.end());

STL去重

首先是2Sum题目，构建一个hash表查找数的对应下标，求两个数的和，找到剩余差去hash查找。

https://leetcode.com/problems/two-sum/

 1 class Solution { 2 public: 3     vector<int> twoSum(vector<int>& nums, int target) { 4         unordered_map <int,int> mapping; 5         vector<int> result; 6         for(int i=0;i<nums.size();i++) 7             mapping[nums[i]]=i; 8         for(int i=0;i<nums.size();i++) 9         {10             const int gap=target-nums[i];11             if(mapping.find(gap)!=mapping.end() && mapping[gap]>i)12             {13                 result.push_back(i+1);14                 result.push_back(mapping[gap]+1);15             }16         }17         return result;18     }19 };

2Sum + hash

对于3Sum题目，使用夹逼法，移动两个指针跳过重复的数字。

https://leetcode.com/problems/3sum/

 1 class Solution { 2 public: 3     vector<vector<int> > threeSum(vector<int> &num) { 4         vector<vector<int> > res; 5         std::sort(num.begin(), num.end()); 6         for (int i = 0; i < num.size(); i++) { 7             int target = -num[i]; 8             int front = i + 1; 9             int back = num.size() - 1;10             while (front < back) {11                 int sum = num[front] + num[back];12                 // Finding answer which start from number num[i]13                 if (sum < target)14                     front++;15                 else if (sum > target)16                     back--;17                 else {18                     res.push_back({num[i],num[front],num[back]});19                     // Processing duplicates of Number 220                     // Rolling the front pointer to the next different number forwards21                     front++;22                     while (front < back && num[front] == num[front-1]) front++;23                     // Processing duplicates of Number 324                     // Rolling the back pointer to the next different number backwards25                     back--;26                     while (front < back && num[back] == num[back+1]) back--;27                 }28             }29             // Processing duplicates of Number 130             while (i + 1 < num.size() && num[i + 1] == num[i]) i++;31         }32         return res;33     }34 };

3Sum + double sides approximate

对于4Sum题目，首先可以使用夹逼法，可以看出这个可以推广到NSum，另外可以结合hash进行搜索。

https://leetcode.com/problems/4sum/

 1 class Solution { 2 public: 3     vector<vector<int> > fourSum(vector<int> &num, int target) { 4         vector<vector<int> > res; 5         if (num.empty()) 6             return res; 7         std::sort(num.begin(),num.end()); 8         for(int i=0;i<num.size();i++) 9         {10             int target_3 = target - num[i];11             for(int j=i+1;j<num.size();j++)12             {13                 int target_2 = target_3 - num[j];14                 int front = j+1;15                 int back = num.size() - 1;16                 17                 while(front < back)18                 {19                     int two_sum=num[front]+num[back];20                     if(two_sum < target_2){21                         front ++;22                     }23                     else if (two_sum > target_2){24                         back --;25                     }26                     else{27                         res.push_back({num[i],num[j],num[front],num[back]});28                         front++;29                         back--;30                         // Processing the duplicates of number 331                         while(front < back && num[front]==num[front-1]) front++;32                         // Processing the duplicates of number 433                         while(front < back && num[back]==num[back+1]) back--; 34                     }35                 }36                 // Processing the duplicates of number 237                 while(j + 1 < num.size() && num[j + 1] == num[j]) ++j;38             }39             // Processing the duplicates of number 140             while(i + 1 < num.size() && num[i + 1] == num[i]) ++i;41         }42         return res;43     }44 };

4Sum-Double Sides Approximate

 1 class Solution { 2 public: 3     vector<vector<int>> fourSum(vector<int>& num, int target) { 4         vector<vector<int>> result; 5         if (num.size() < 4) return result; 6         sort(num.begin(), num.end()); 7  8         unordered_map<int,vector<pair<int,int> > > cache; 9         10         for(int a=0;a<num.size();a++)11         {12             for(int b=a+1;b<num.size();b++)13             {14                 cache[num[a]+num[b]].push_back(pair<int,int>(a,b));15             }16         }17         18         for(int c=0;c<num.size();c++)19         {20             for(int d=c+1;d<num.size();d++)21             {22                 int key=target-num[c]-num[d];23                 if(cache.find(key) == cache.end()) continue;24                 const auto& vec = cache[key];25                 for(int k=0;k<vec.size();++k)26                 {27                     if(c<=vec[k].second)28                     {29                         continue;30                     }31                     result.push_back( {num[vec[k].first],num[vec[k].second],num[c],num[d]} );32                 }33             }34         }35         sort(result.begin(), result.end());36         result.erase(unique(result.begin(), result.end()), result.end());37         return result;38     }39 };

4Sum-hash+double sides approximate

参考解答

灵魂机器leetcode解答https://github.com/soulmachine/leetcode

https://leetcode.com/discuss/27198/solution-explanation-comparison-problem-easy-understand

https://leetcode.com/discuss/23595/share-my-solution-around-50ms-with-explanation-and-comments

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