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:
For example, given candidate set 2,3,6,7 and target 7,
A solution set is:
[7]
[2, 2, 3]
给定一个候选数集合candidates，和一个目标值target。从候选数集合中选出所有可能的组合，使得它们的和为target。候选集中的数可重复选择，没有次数限制。
几点说明：
1. 本题所有的数都是正数
2. 组合中的数非递减排列
3. 组合不能重复
class Solution {
public:
void dfs(vector<vector<int> >&result, vector<int>&candidates, int&target, vector<int>combination, int sum, int startIndex){
//sum-组合中已有数的和
//startIndex-选择组合中下个数的起始位（即从候选数集中的startIndex索引位上的数开始选择）
if(sum==target)result.push_back(combination);
else if(sum<target){
for(int i=startIndex; i<candidates.size(); i++){
combination.push_back(candidates[i]);
dfs(result,candidates,target,combination,sum+candidates[i],i);
combination.pop_back();
}
}
}
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
vector<vector<int> > result;
int size=candidates.size();
if(size==0)return result;
// 排序
sort(candidates.begin(), candidates.end());
vector<int>combination;
dfs(result, candidates, target, combination, 0, 0);
return result;
}
};LeetCode: Combination Sum [038],布布扣,bubuko.com
LeetCode: Combination Sum [038]
原文地址：http://blog.csdn.net/harryhuang1990/article/details/26336835
