[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。

【代码】

/*********************************
*   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;
        }//if
        // 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;
        }//if
        for(int i = start;i < len;++i){
            // 剪枝
            if(target < candidates[i]){
                return;
            }//if
            path.push_back(candidates[i]);
            DFS(candidates,target-candidates[i],i,path,result);
            path.pop_back();
        }//for
    }
};

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;
}