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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] ]
采用深度优先搜索的思想，对于输入candidates=[1,2] ，target=3，遍历的方向如图：
1 #include "stdafx.h" 2 3 #include <iostream> 4 #include <vector> 5 #include <algorithm> 6 7 using namespace std; 8 void helper(int pos, int base, int target, vector<int>& candidates, vector<int>& path, vector<vector<int> >& result) 9 { 10 if (base > target) 11 return; 12 if (base == target) 13 { 14 result.push_back(path); 15 return; 16 } 17 for (int i = pos; i < candidates.size(); ++i) 18 { 19 path.push_back(candidates[i]); 20 helper(i, base + candidates[i], target, candidates, path, result); 21 path.pop_back(); 22 } 23 24 } 25 vector<vector<int> > combinationSum(vector<int>& candidates, int target) 26 { 27 vector<vector<int> > result; 28 vector<int> path; 29 sort(candidates.begin(), candidates.end()); 30 vector<int>::iterator it = unique(candidates.begin(),candidates.end()); 31 candidates.erase(it, candidates.end()); 32 helper(0, 0, target, candidates, path, result); 33 return result; 34 } 35 int main() 36 { 37 vector<vector<int> > result; 38 vector<int> candidates; 39 candidates.push_back(2); 40 candidates.push_back(2); 41 candidates.push_back(3); 42 candidates.push_back(7); 43 int target = 7; 44 result = combinationSum(candidates, target); 45 46 return 0; 47 }
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原文地址：http://www.cnblogs.com/hhboboy/p/5857345.html
