【leetcode刷题笔记】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] 

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题解：类似八皇后问题。每次把candidates[i]加入到numbers列表中，然后递归的搜索target-candidates[i]的组成方法。

要注意的一点是，将candidates[i]加入到numbers列表后，i前面的元素就不能再往numbers里面加了，只能再加入i本身或者i后面的元素。所以在递归函数中要有一个参数start，表明只有start本身或者以后的元素可以加入numbers中。每当递归函数参数target等于0的时候，说明找到了一组答案在numbers中，把numbers加入到answer里面就可以了。

代码如下：

 1 public class Solution {

 2     private void combiDfs(int[] candidates,int target,List<List<Integer>> answer,List<Integer> numbers,int start){

 3         if(target == 0){

 4             answer.add(new ArrayList<Integer>(numbers));

 5             return;

 6         }

 7         for(int i = start;i < candidates.length;i++){

 8             if(candidates[i] > target)

 9                 break;

10             numbers.add(candidates[i]);

11             combiDfs(candidates, target-candidates[i], answer, numbers,i);

12             numbers.remove(numbers.size()-1);

13         }

14     }

15     public List<List<Integer>> combinationSum(int[] candidates, int target) {

16         List<List<Integer>> answer = new ArrayList<List<Integer>>();

17         List<Integer> numbers = new ArrayList<Integer>();

18         Arrays.sort(candidates);

19         combiDfs(candidates, target, answer, numbers,0);

20         

21         return answer;

22         

23     }

24 }