3-sum

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:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
The solution set must not contain duplicate triplets.

在一个数组中 ，num = -1 ，-2 ，1，2
找出 a b c in num ，并且 a+b+c = 0;方案的个数，并且打印出方案个数，（每个方案直接不重复）
最开始的代码 复杂度接近 o(n^3) ,数据一大就超时了

 #include <iostream>#include <vector>#include <algorithm>#include <set>using namespace std;class Solution {public:    vector<vector<int> > threeSum(vector<int> &num) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        vector<vector <int > >re;        set<vector <int >  > reSet;        sort(num.begin() ,num.end() );        int len = num.size();        for(int i = 0 ; i < len ; i ++){            for(int j = i + 1 ; j < len ; j ++){                int t = num[i] + num[j];                if(t > 0 ) break;                for(int k = j + 1 ; k < len ; k ++){                    int sum = t + num[k];                    if(sum == 0  ) {                        vector<int> r;                        r.push_back(num[i]);                        r.push_back(num[j]);                        r.push_back(num[k]);                        if(reSet.find(r) == reSet.end()){                            re.push_back(r);                            reSet.insert(r);                        }                    }                    else if(sum > 0 ) break;                }            }        }        return re;    }};

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).

写代码 不要想当然说 输入数据的范围，要考虑各种情况。之前 int delta = 0x3fffffff; 竟然写成 = 3*target 如果target =0 ，结果就很多错误

 #include<iostream>#include<vector>#include<algorithm>#include<map>using namespace::std;class Solution {public:    int threeSumClosest(vector<int> &num, int target) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        sort(num.begin(),num.end());        int sum = 0;        int re = 0;        int delta = 0x3fffffff;        int len = num.size();        for(int i = 0 ; i < len - 2 ; i ++){            int l = i +1 ;            int r = len -1;            int pre = -1;            int now = 1;            while(l < r ){                sum = num[i] + num[l] + num[r];                if(sum == target) return target;                else if(sum < target ){                    if(abs(sum - target) < delta) delta = abs(sum - target),re = sum;                    l ++ ;                }                else{                    if(abs(sum - target) < delta) delta = abs(sum - target) ,re = sum;                    r --;                }            }        }//end for        return re;    }};

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