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.
• 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] 

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思路： 和 combination sum 一样， 用DFS ， 但是每个元素只能用一次， 为了没有同值的不同元素重复使用， 设置一个 pre  来存储上一个元素的值， 如果相同， 直接跳到下一个元素。

public class Solution {    public List<List<Integer>> combinationSum2(int[] num, int target) {        List<List<Integer>> ret = new ArrayList<List<Integer>>();        Arrays.sort(num);        List<Integer> sol = new ArrayList<Integer>();        dfs(num, 0, 0, target, sol, ret);        return ret;    }        private void dfs(int[] nums, int start, int sum, int target, List<Integer> sol,  List<List<Integer>> ret){        if(sum == target){            ret.add(new ArrayList<Integer>(sol));            return;        }else if(sum > target){            return;        }        int pre = -1;        for(int i = start; i < nums.length; i++){            if(nums[i] != pre){                sol.add(nums[i]);                dfs(nums, i + 1, sum + nums[i], target, sol, ret);                sol.remove(sol.size() - 1);                pre = nums[i];            }        }    }}