LeetCode——039_星野明博客-优快云博客

本文链接：https://blog.youkuaiyun.com/langmanqishizaijia/article/details/51171986

本文介绍了一道典型的深度遍历算法题目——组合总和问题。该问题要求在给定候选数集合与目标数的情况下，找出所有可能的组合，使得候选数之和等于目标数。文章详细阐述了解决方案，包括对数组进行排序、使用深度优先搜索策略以及如何避免重复组合。

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/*
39. Combination Sum My Submissions QuestionEditorial Solution
Total Accepted: 88140 Total Submissions: 286018 Difficulty: Medium
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:
All numbers (including target) will be positive integers.
Elements in a combination (a1, a2, … , ak) must be in non-descending order. (ie, a1 ≤ a2 ≤ … ≤ ak).
The solution set must not contain duplicate combinations.
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
*/
/*
解题思路：
深度遍历的典型题目，可以重复使用同一个元素，所以每次往下层递归的时候游标并不+1，还是i.
注意事项：
首先要对数组进行排序，只有这样才能保证结果都是有序的
防止重复，暂时的结果先放在一个set中
*/

class Solution {
public:
    vector<vector<int>> combinationSum(vector<int>& candidates, int target) {

        //本题属于深度遍历

        if(candidates.size()==0)return {};
        set<vector<int>> _set;
        vector<int> out;
        sort(candidates.begin(),candidates.end());
        dfs(candidates,0,target,out,_set);
        return vector<vector<int>> (_set.begin(),_set.end());
    }

    void dfs(vector<int>& candidates,int start,int target,vector<int>&out,set<vector<int>>&_set){

        if(target==0){
            _set.insert(out);
            return ;
        }
        if(target<0)return ;
        for(int i=start;i<candidates.size();i++){
            out.push_back(candidates[i]);
            dfs(candidates,i,target-out.back(),out,_set);
            out.pop_back();
        }

    }
};