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:
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)

和3Sum思想一样，这个复杂度是O(n^3)

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        vector<vector<int> > res;        if (num.empty())            return res;        std::sort(num.begin(),num.end());        for (int i = 0; i < num.size(); i++) {            int target_3 = target - num[i];            for (int j = i + 1; j < num.size(); j++) {                int target_2 = target_3 - num[j];                int front = j + 1;                int back = num.size() - 1;                while(front < back) {                    int two_sum = num[front] + num[back];                    if (two_sum < target_2) front++;                    else if (two_sum > target_2) back--;                    else {                        vector<int> quadruplet(4, 0);                        quadruplet[0] = num[i];                        quadruplet[1] = num[j];                        quadruplet[2] = num[front];                        quadruplet[3] = num[back];                        res.push_back(quadruplet);                        // Processing the duplicates of number 3                        while (front < back && num[front] == quadruplet[2]) ++front;                        // Processing the duplicates of number 4                        while (front < back && num[back] == quadruplet[3]) --back;                    }                }                // Processing the duplicates of number 2                while(j + 1 < num.size() && num[j + 1] == num[j]) ++j;            }            // Processing the duplicates of number 1            while (i + 1 < num.size() && num[i + 1] == num[i]) ++i;        }        return res;    }};