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 
#include 
#include 
#include 
using namespace std;


class Solution {
public:
    vector > threeSum(vector &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector >re;

        set  > 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 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
#include
#include
#include

原始博客地址