【LeetCode】leetcode sum问题汇总小结

1. 2Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,

return [0, 1].

两个数相加，所以通过hash表记录数a与数target-a有没有出现过即可，因为还要记录下标，所以用hash。

并且由于从前到后遍历，且只有一个答案，所以遍历一次即可。

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {         vector<int> rst;         unordered_map<int,int> hash;         for(int i=0;i<nums.size();i++)         {         int n=target-nums[i];             if(hash.find(n)!=hash.end())             {                 rst.push_back(hash[target-nums[i]]);                 rst.push_back(i);                 return rst;             }else hash[nums[i]]=i;         }         return rst;    }};

2. 3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],A solution set is:[  [-1, 0, 1],  [-1, -1, 2]]

这题结果可能有重复，并且不要求输出下标，所以可以先排序，再每次固定第一个数，对后两个数用two pointer。front back

如果front+back+first>0，back--,反之front++.为了避免重复，跳过重复的数。

class Solution {public:    vector<vector<int>> threeSum(vector<int>& nums) {        vector<vector<int>> rst;        sort(nums.begin(),nums.end());        for(int i=0;i<nums.size();i++)        {        if(i>0&&nums[i]==nums[i-1]) continue;   //跳过跟n1重复的            int n1=nums[i];            int front=i+1;            int back=nums.size()-1;            while(front<back)            {                n2=nums[front];                n3=nums[back];                if(n1+n2+n3<0) front++;                else if(n1+n2+n3>0) back--;                else{                vector<int> tmp{n1,n2,n3};                rst.push_back(tmp);                while(front<back&&nums[front]==n2) front++;  //跳过跟n2重复的                while(front<back&&nums[back]==n3) back--;    //跳过跟n3重复的                }            }        }                       return rst;    }};

3. 3Sum Closest

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

    For example, given array S = {-1 2 1 -4}, and target = 1.    The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

这题跟上题类似，不过答案只有一个，且每次要判断是不是最近的点，求什么设什么。

class Solution {public:    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {        int rst=0,len=A.size();        unordered_map<int,int> hasha,hashb,hashc,hashd;        sort(A.begin(),A.end());        sort(B.begin(),B.end());        sort(C.begin(),C.end());        sort(D.begin(),D.end());        inithash(A,hasha);        inithash(B,hashb);        inithash(C,hashc);        inithash(D,hashd);        if(A[0]+B[0]+C[0]+D[0]>0) return 0;        if(A[len-1]+B[len-1]+C[len-1]+D[len-1]<0) return 0;        for(int i=0;i<len;i++)        {            if(i>0&&A[i]==A[i-1]) continue;            int sum3=0-A[i];            for(int j=0;j<len;j++){                if(j>0&&A[j]==A[j-1]) continue;                if(B[j]+C[j]+D[j]>0) break;                if(B[len-1]+B[len-1]+C[len-1]<0)                            }        }                            }    void inithash(vector<int>&A, unordered_map<int,int> &hash){        for(int i=0;i<A.size();i++){            if(hash.find(a)==hash.end()) hash[a]=1;            else hash[a]++;        }    }};

18. 4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.A solution set is:[  [-1,  0, 0, 1],  [-2, -1, 1, 2],  [-2,  0, 0, 2]]

跟3sum比较像，不过升级到四个 可以固定前两个，再加个循环，O(n^3)的复杂度。参考大神的代码发现还可以剪枝

class Solution {public:    vector<vector<int>> fourSum(vector<int>& nums, int target) {        vector<vector<int>> total;        int n = nums.size();        if(n<4)  return total;        sort(nums.begin(),nums.end());        for(int i=0;i<n-3;i++)        {            if(i>0&&nums[i]==nums[i-1]) continue;            if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;            if(nums[i]+nums[n-3]+nums[n-2]+nums[n-1]<target) continue;            for(int j=i+1;j<n-2;j++)            {                if(j>i+1&&nums[j]==nums[j-1]) continue;                if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;                if(nums[i]+nums[j]+nums[n-2]+nums[n-1]<target) continue;                int left=j+1,right=n-1;                while(left<right){                    int sum=nums[left]+nums[right]+nums[i]+nums[j];                    if(sum<target) left++;                    else if(sum>target) right--;                    else{                        total.push_back(vector<int>{nums[i],nums[j],nums[left],nums[right]});                        do{left++;}while(nums[left]==nums[left-1]&&left<right);                        do{right--;}while(nums[right]==nums[right+1]&&left<right);                    }                }            }        }        return total;    }};

19. 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

因为这题只需要求个数，而不需要具体的下标值，所以可以两个一组求和，对各和的值做hash记录，遍历一个hash，在另一个hash中查找是否出现过使和为target的值，两个值的个数求和。

class Solution {public:    void fillMap(vector<int>& A, vector<int>& B, unordered_map<int,int> &m)    {        int n = A.size();        for(int i = 0; i < n; ++i)        for(int j = 0; j < n; ++j)          ++m[A[i] + B[j]];              }    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {        unordered_map<int,int> m1, m2;        fillMap(A, B, m1);        fillMap(C, D, m2);        int res = 0;        for(auto it = m1.begin(); it != m1.end(); ++it)        {           auto it2 = m2.find(-it->first);           if(it2 != m2.end())             res += it->second*it2->second;        }        return res;    }