leetcode -- 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, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
• 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] 

本题是求解所有可能的组合数，可用DFS来求解，如需求解最优解，需使用DP

DFS的递归函数中只处理当前状态节点n，而不关心它的下一状态

 1 public class Solution {
 2     public ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int len = candidates.length;
 6         ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
 7         if(len == 0)
 8             return results;
 9         
10         Arrays.sort(candidates);
11         
12         ArrayList<Integer> result = new ArrayList<Integer>();
13         int level = 0;
14         int sum = 0;
15         DFS(candidates, level, sum, target, result, results);
16         return results;
17     }
18     
19     public void DFS(int[] candidates, int level, int sum, int target, ArrayList<Integer> result,
20         ArrayList<ArrayList<Integer>> results){
21         if(sum > target)
22             return;
23         
24         if(sum == target){
25             ArrayList<Integer> tmp = new ArrayList<Integer>();
26             tmp.addAll(result);
27             results.add(tmp);
28             return;
29         }
30         if(sum < target){
31             for(int i = level; i < candidates.length; i++){
32                 sum += candidates[i];
33                 result.add(candidates[i]);
34                 DFS(candidates, i, sum, target, result, results);
35                 result.remove(result.size() - 1);
36                 sum -= candidates[i];
37             }
38         }
39     }
40 }

ref：

DFS 算法入门：http://rapheal.iteye.com/blog/1526863