[Leetcode] 4 Sum [Unsolved]-优快云博客

本文探讨了如何寻找数组中四个元素的组合，使它们的和等于特定目标值的问题。利用深度优先搜索（DFS）结合两数之和的方法，确保找到所有独特的四元组，并按非递减顺序排列，同时避免重复解决方案。

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 â‰¤ b â‰¤ c â‰¤ 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)

DFS！用一个 Used Vector 避免重复。

大集合没过，目测应该用 DFS + Two Sum 的方法。

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > solution;
        vector<int> answer;
        vector<bool> used (num.size(),false);
        sort(num.begin(),num.end());
        dfs(used,solution,answer,num,0,0,target);
        return solution;
        
    }
    
    void dfs (vector<bool> &used, vector<vector<int> > &solution, vector<int> &answer, vector<int> &num, int depth, int index, int target)
    {
        if(depth==4)
        {
            if(target==0)
            {
                solution.push_back(answer);
            }
            return;
        }
        
        for(int i= index; i<num.size(); i++)
        {
            
            if(i!=0 && num[i]==num[i-1] && used[i-1]==false)
            {
                
            }
            else
            {
                answer.push_back(num[i]);
                used[i] = true;
                dfs(used,solution,answer,num,depth+1,i+1,target-num[i]);
                answer.pop_back();
                used[i] = false;                
            }            
        } 
    }
};