[LeetCode]39.Combination Sum

【题目】

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] 

【分析】

基本思路是先排好序，然后每次递归中把剩下的元素一一加到结果集合中，并且把目标减去加入的元素，然后把剩下元素（包括当前加入的元素）放到下一层递归中解决子问题。以start记录我们选到了第几个值，并且一直往后选，这样可以避免选到重复的子集。

/**********************************   日期：2015-01-27*   作者：SJF0115*   题目: 39.Combination Sum*   网址：https://oj.leetcode.com/problems/combination-sum/*   结果：AC*   来源：LeetCode*   博客：**********************************/#include <iostream>#include <vector>#include <algorithm>using namespace std;class Solution {public:    vector<vector<int> > combinationSum(vector<int> &candidates, int target) {        // 中间结果        vector<int> path;        // 最终结果        vector<vector<int> > result;        int size = candidates.size();        if(size <= 0){            return result;        }        // 排序        sort(candidates.begin(),candidates.end());        // 递归        DFS(candidates,target,0,path,result);        return result;    }private:    void DFS(vector<int> &candidates, int target,int start,vector<int> &path,vector<vector<int> > &result){        int len = candidates.size();        // 找到一组组合和为target        if(target == 0){            result.push_back(path);            return;        }        for(int i = start;i < len;++i){            // 剪枝            if(target < candidates[i]){                return;            }            path.push_back(candidates[i]);            DFS(candidates,target-candidates[i],i,path,result);            path.pop_back();        }    }};int main(){    Solution solution;    int target = 7;    vector<int> vec;    vec.push_back(2);    vec.push_back(3);    vec.push_back(6);    vec.push_back(7);    vector<vector<int> > result = solution.combinationSum(vec,target);    // 输出    for(int i = 0;i < result.size();++i){        for(int j = 0;j < result[i].size();++j){            cout<<result[i][j]<<" ";        }        cout<<endl;    }    return 0;}

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