problem：
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]
thinking：
（1）这道题的难点是怎么处理每个数可以重复出现
（2）采取DFS策略，遇到符合要求的解往下一层继续搜索，遇到不合适的则回溯搜索。
plz 理解回溯偶的思想，会构造和使用DFS！！！
code：
class Solution {
private:
vector<vector<int> > ret;
vector<int> a;//记录有效解
public:
void solve(int dep, int maxDep, int target, vector<int> &cand)
{
if (target < 0) //回溯条件
return;
if (dep == maxDep) //DFS终止条件
{
if (target == 0) //获取有效解
{
vector<int> res;
for(int i = 0; i < maxDep; i++)
for(int j = 0; j < a[i]; j++)
res.push_back(cand[i]);
ret.push_back(res);
}
return;
}
for(int i = 0; i <= target / cand[dep]; i++)
{
a[dep] = i;
solve(dep + 1, maxDep, target - cand[dep] * i, cand);
}
}
vector<vector<int> > combinationSum(vector<int> &candidates, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
sort(candidates.begin(), candidates.end());
a.resize(candidates.size());
ret.clear();
if (candidates.size() == 0)
return ret;
solve(0, candidates.size(), target, candidates);
return ret;
}
};
leetcode || 39、Combination Sum
原文：http://blog.csdn.net/hustyangju/article/details/44649889