LintCode(135) Combination Sum

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.

For example, given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

Have you met this question in a real interview? 

Yes

Example

given candidate set 2,3,6,7 and target 7, 
A solution set is: 
[7] 
[2, 2, 3] 

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.

这基本算是回溯法的经典题目了，之前朋友面试碰到过类似的，就是给 1， 5， 10， 25的硬币，求有多少种组合可以组成一个给定的币值， 其实就是这个题目。

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class Solution { public: /** * @param candidates: A list of integers * @param target:An integer * @return: A list of lists of integers */ vector<vector<int> > combinationSum(vector<int> &candidates, int target) { // write your code here vector<vector<int>> result; vector<int> solution; sort(candidates.begin(), candidates.end()); helper(result, solution, candidates, 0, target); return result; } void helper(vector<vector<int>>& result, vector<int>& solution, vector<int>& candidates, int ind, int target){ if (target<0) return; if (target==0){ result.push_back(solution); return; } for (int i=ind; i<candidates.size(); i++){ solution.push_back(candidates[i]); helper(result,solution, candidates, i, target-candidates[i]); solution.pop_back(); } } };