【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 ( a 1, a 2, … , a k) must be in non-descending order. (ie, a 1 ≤ a 2 ≤ … ≤ 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]

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：

 public class Solution {

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

         if(target == 0){

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

             return;

         }

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

             if(candidates[i] > target)

                 break;

             numbers.add(candidates[i]);

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

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

         }

     }

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

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

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

         Arrays.sort(candidates);

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

         return answer;

     }

 }