Leetcode 4Sum

先来一个暴力解：

class Solution {

public:

    vector<vector<int> > fourSum(vector<int> &num, int target) {

        vector<vector<int> > res;

        vector<int> r;

        r.resize(4);

        sort(num.begin(), num.end());



        int len = num.size();

        for (int i=0; i<len; i++) {

            if (i!=0 && num[i] == num[i-1]) continue;   // skip dup

            for (int j=i+1; j<len; j++) {

                int psum = num[i] + num[j];

                if (num[j] >= 0 && psum > target) break;

                if (j!=i+1 && num[j] == num[j-1]) continue; // skip dup

                r[0] = num[i], r[1] = num[j];

                

                int ptarget = target - psum;



                int p = j+1;

                int q = len - 1;

                while (p < q) {

                    if (num[p] >= 0 && num[p] > ptarget) break;

                    if (num[q] < 0 && num[q] < ptarget) break;

                    if (p != j+1 && num[p] == num[p-1]) {   // skip dup

                        p++;

                        continue;

                    }

                    if (q != len-1 && num[q] == num[q+1]) { // skip dup

                        q--;

                        continue;

                    }

                    int s = num[p] + num[q];

                    if (s < ptarget) {

                        p++;

                    } else if (s > ptarget) {

                        q--;

                    } else {

                        r[2] = num[p];

                        r[3] = num[q];

                        res.push_back(r);

                        p++, q--;

                    }

                }

            }

        }

        return res;

    }

};

有些快速跳出的语句加了反而更慢。

第二轮：

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)

class Solution {

public:

    vector<vector<int> > fourSum(vector<int> &num, int target) {

        sort(num.begin(), num.end());

        

        vector<vector<int> > res;

        int len = num.size();

        if (len < 4) {

            res;

        }

        

        for (int i=0; i<len; i++) {

            if (i!=0 && num[i] == num[i-1]) continue;

            for (int j=i+1; j<len; j++) {

                if (j!=i+1 && num[j] == num[j-1]) continue; 

                int p = j+1;

                int q = len -1;

                int rest = target - num[i] - num[j];

                

                while (p < q) {

                    int t = num[p] + num[q];

                    if (t == rest) {

                        if (!res.empty() 

                            && res.back()[0] == num[i] 

                            && res.back()[1] == num[j] 

                            && res.back()[2] == num[p]

                            && res.back()[3] == num[q]) {

                        } else {

                            res.push_back(vector<int>({num[i], num[j], num[p], num[q]}));

                        }

                        p++;

                    } else if (t > rest) {

                        q--;

                    } else {

                        p++;

                    }

                }

            }

        }

        return res;

    }

};