40. Combination Sum II

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

• All numbers (including target) will be positive integers.
• 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]
]

本题让我对这一系列题目又有了新的感觉，先说区别，本题前面计算过的数组元素，后面的元素计算的时候就不需要回来把之前的元素加上了，也就是说，相当于小卒一去不回头的感觉，它和combination sum1的区别最大的就是会不会包含它本身。代码如下：

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(candidates);
        backtracking(res,candidates,target,new ArrayList<Integer>(),0);
        return res;
    }
    public void backtracking(List<List<Integer>> res,int[] candidates,int target,List<Integer> list,int start){
        if(target==0){
            res.add(new ArrayList<Integer>(list));
        }else if(target<0) return;
        else{
            for(int i=start;i<candidates.length;i++){
                if(i>start&&candidates[i]==candidates[i-1]){
                    continue;
                }
                list.add(candidates[i]);
                backtracking(res,candidates,target-candidates[i],list,i+1);
                list.remove(list.size()-1);
            }
        }
    }
}