leetcode-Combination Sum-39

输入一个数组和target，选择任意个数的元素，求和为target的组合，每个元素可以选择多次

dfs，回溯，因为每个元素可以选择多次，所以向下搜索的时候从当前元素开始

同类题：http://blog.csdn.net/AC_0_summer/article/details/48293581

 1 class Solution {
 2 public:
 3     void dfs(vector<vector<int> > &v,vector<int> vv,vector<int> a,int i,int target,int sum){
 4         if(sum>target) return;
 5         if(sum==target){
 6             v.push_back(vv);
 7             return;
 8         }
 9        // vv.push_back(a[i]);
10         for(int j=i;j<a.size();j++){
11             if(sum+a[j]<=target){
12                 vv.push_back(a[j]);
13                 dfs(v,vv,a,j,target,sum+a[j]);
14                 vv.pop_back();
15             }
16         }
17     }
18     vector<vector<int> > combinationSum(vector<int>& candidates, int target) {
19         vector<vector<int> > v;
20         vector<int> vv;
21         int sum=0;
22         sort(candidates.begin(),candidates.end());
23         for(int i=0;i<candidates.size();i++){
24             if(candidates[i]<=target){
25                 vv.push_back(candidates[i]);
26                 dfs(v,vv,candidates,i,target,sum+candidates[i]);
27                 vv.pop_back();
28             }
29         }
30         if(v.size()==0) return v;
31         vector<vector<int> > ans;
32         ans.push_back(v[0]);
33         int k=0;
34         for(int i=1;i<v.size();i++){
35             if(ans[k].size()!=v[i].size()){
36                 ans.push_back(v[i]);
37                 k++;
38                 continue;
39             }
40             int ok=0;
41             for(int j=0;j<ans[k].size();j++){
42                 if(v[i][j]!=ans[k][j]){
43                     ok=1;break;
44                 }
45             }
46             if(ok){
47                 ans.push_back(v[i]);
48                 k++;
49             }
50         }
51         return ans;
52     }
53 };