ThinkChat🤖让你学习和工作更高效,注册即送10W Token,即刻开启你的AI之旅 广告
**一. 题目描述** 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. (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] ~~~ **二. 题目分析** 题目大意是:有一个正整数集合C和一个正整数目标T。现从C中选出一些数,使其累加和恰好等于T(C中的每个数都可以取若干次),求所有不同的取数方案,是一道典型的深度优先搜索题。 题目的一个细节是,正整数集合C可能存在**重复的数**,而这些数用于组合时,会产生重复的排列组合,如下图所示,而题目要求:The solution set must not contain duplicate combinations. ![](https://box.kancloud.cn/2016-01-05_568bb5f11dc9e.jpg) 我的解决方法是对正整数集合C先进行从小到大排序,然后使用vector的erase方法去除集合C的重复元素,然后再进行DFS的相关运算。 **三. 示例代码** ~~~ #include <iostream> #include <vector> #include <algorithm> using namespace std; class Solution { public: vector<vector<int> > combinationSum(vector<int> &candidates, int target) { vector<int> temp; // 用于存放临时组合 sort(candidates.begin(), candidates.end()); // 结果要求输出的组合不能重复,需要去除candidates中的重复元素 candidates.erase(unique(candidates.begin(), candidates.end()), candidates.end()); combinationDFS(candidates, temp, 0, target); return result; } private: vector<vector<int> > result; void combinationDFS(vector<int> &candidates, vector<int> &temp, size_t index, int target) { if (target == 0) // 得到满足目标的一组解 { result.push_back(temp); return; } else { for (size_t i = index; i < candidates.size(); ++i) { if (candidates[i] > target) return; temp.push_back(candidates[i]); combinationDFS(candidates, temp, i, target - candidates[i]); temp.pop_back(); } } } }; ~~~ ![](https://box.kancloud.cn/2016-01-05_568bb5f136576.jpg) **四. 小结** 该题是一道经典的DFS题目,后续还有Combination Sum II和Combination Sum III,需要深入研究。