【Leetcode】【Medium】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.

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.

 

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

 

解题思路:

使用回溯的思想穷举可能的结果。

代码:

 1 class Solution {
 2 public:
 3     vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
 4         vector<vector<int>> lst;
 5         vector<int> ans;
 6         sort(candidates.begin(), candidates.end());
 7         Backtracking(lst, ans, candidates, 0, target);
 8         return lst;
 9     }
10     
11     void Backtracking(vector<vector<int>> &lst, vector<int> ans, vector<int> candidates, int idx, int left) {
12         for (int i = idx; i < candidates.size(); ++i) {
13             int cur_left = left - candidates[i];
14             if (cur_left == 0) {
15                 ans.push_back(candidates[i]);
16                 lst.push_back(ans);
17                 return;
18             }
19             if (cur_left > 0) {
20                 ans.push_back(candidates[i]);
21                 Backtracking(lst, ans, candidates, i, cur_left);
22                 ans.pop_back();
23             } else {
24                 return;
25             } 
26         }
27     }
28 };

 

另:是否还有DP/DFS等其他思路。

posted @ 2015-03-06 04:44  胡潇  阅读(152)  评论(0编辑  收藏  举报