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:
- 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 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：
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;
}
int prev = -1;
for(int i = start;i < candidates.length;i++){
if(candidates[i] > target)
break;
if(prev != -1 && prev == candidates[i])
continue;
numbers.add(candidates[i]);
combiDfs(candidates, target-candidates[i], answer, numbers,i+1);
numbers.remove(numbers.size()-1);
prev = candidates[i];
}
}
public List<List<Integer>> combinationSum2(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;
}
}