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:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• 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)

与3Sum很相似的一道题，复杂度成了n³logn。但是如果直接这样来的话，TLE，我也是ac以后尝试才发现的。。

其实，就是要加剪枝了。就是如果给定当前的3个数字num[i],num[j],num[h]，其中i<j<h，如果num[h+1]>(target-num[i]-num[j]-num[h])，则中断循环。

对于第一层和第二层循环也可以进行相似的判断。

class Solution {public:    bool find(vector<int> &num,int start,int end,int target){        while(start<=end){            int mid = (start+end)/2;            if(num[mid]>target){                end = mid-1;            }else if(num[mid]==target){                return true;            }else{                start = mid+1;            }        }        return false;    }    vector<vector<int> > fourSum(vector<int> &num, int target) {        sort(num.begin(),num.end());        vector<vector<int> > vv;        vector<int> v;        int n = num.size();        for(int i=0;i<n-3;){            v.push_back(num[i]);            int t1 = target-num[i];            if(t1<(num[i+1]+num[i+2]+num[i+3]))                break;            for(int j=i+1;j<n-2;){                int t2 = target-num[i]-num[j];                if(t2<num[j+1]+num[j+2])                    break;                v.push_back(num[j]);                for(int h = j+1;h<n-1;){                    int tmp = target-num[i]-num[j]-num[h];                    if(h+1<n && num[h+1]>tmp)                        break;                    if(find(num,h+1,n-1,target-num[i]-num[j]-num[h])){                        v.push_back(num[h]);                        v.push_back(target-num[i]-num[j]-num[h]);                        vv.push_back(v);                        v.pop_back();                        v.pop_back();                    }                    h++;                    while(h<n && num[h]==num[h-1])                        h++;                }                j++;                while(j<n && num[j]==num[j-1])                    j++;                v.pop_back();            }            i++;            while(i<n && num[i]==num[i-1])                i++;            v.pop_back();        }        return vv;    }};