Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums toT.
Each number in C may only be used once in the combination.
Note:
For example, given candidate set 10,1,2,7,6,1,5
and target 8
,
A solution set is: [1, 7]
[1, 2, 5]
[2, 6]
[1, 1, 6]
跟Combination Sum的差别就是上面红色的那行字,只要把递归时候的起点由i改为i+1即可,参见下列代码中高亮的部分:
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 8 int prev = -1; 9 10 for(int i = start;i < candidates.length;i++){ 11 if(candidates[i] > target) 12 break; 13 if(prev != -1 && prev == candidates[i]) 14 continue; 15 16 numbers.add(candidates[i]); 17 combiDfs(candidates, target-candidates[i], answer, numbers,i+1); 18 numbers.remove(numbers.size()-1); 19 20 prev = candidates[i]; 21 } 22 } 23 public List<List<Integer>> combinationSum2(int[] candidates, int target) { 24 List<List<Integer>> answer = new ArrayList<List<Integer>>(); 25 List<Integer> numbers = new ArrayList<Integer>(); 26 Arrays.sort(candidates); 27 combiDfs(candidates, target, answer, numbers,0); 28 29 return answer; 30 31 } 32 }