LeetCode 113 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]

分析：

这跟Conbination Sum 不同的地方有两处：

1，在DFS递归的时候，因为i位置元素只能用一次，所以递归要 i+1;

2，在判断continue的时候，是 i > start && num[i] == num[i-1]，这个保证 num[start] == num[start-1]的时候也是有效地（即例子中 1,1,6）的情况。

public class Solution {
    public List<List<Integer>> combinationSum2(int[] num, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if(num==null || num.length==0)
            return res;
        Arrays.sort(num);
        dfs(num, 0, target, new ArrayList<Integer>(), res);
        return res;
    }
    
    public void dfs(int[] num, int start, int target, List<Integer> item, List<List<Integer>> res){
        if(target<=0){
            if(target==0)
                res.add(new ArrayList<Integer>(item));
            return;
        }
        for(int i=start; i<num.length; i++){
            //i>start保证num[start]==num[start-1]的时候也可以处理到
            if(i>start && num[i]==num[i-1]) continue;
            item.add(num[i]);
            //每次i+1,保证i位置只使用一次
            dfs(num, i+1, target-num[i], item, res);
            item.remove(item.size()-1);
        }
    }
}