[Lintcode] Combination Sum II

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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.

Example

Given candidate set [10,1,6,7,2,1,5] and target 8,

A solution set is:

[  [1,7],  [1,2,5],  [2,6],  [1,1,6]]

递归即可。考虑去重。

public class Solution {    /**     * @param num: Given the candidate numbers     * @param target: Given the target number     * @return: All the combinations that sum to target     */    public List<List<Integer>> combinationSum2(int[] num, int target) {        List<List<Integer>> res = new ArrayList<List<Integer>>();        Arrays.sort(num);        helper(res, new ArrayList<Integer>(), num, target, 0);        return res;    }        void helper(List<List<Integer>> res, List<Integer> list, int[] num, int target, int index) {        if(target == 0) {            res.add(new ArrayList(list));            return;        }                if(index >= num.length || target < 0) return;                for(int i = index; i < num.length; i++) {            list.add(num[i]);            helper(res, list, num, target - num[i], i + 1);//i + 1 instead of index + 1            while(i + 1 < num.length && num[i] == num[i + 1]) i += 1;//在helper之后去重            //如果在helper之前去重的话，会漏掉结果集中存在重复元素的情况            // 1 1 2 5 6 7 10 -> 1 1 2 OK 在helper前去重漏掉的情况            // 1 1 2 5 6 7 10 -> 1 2 5, 1 2 5 KO 需要去重的情况            list.remove(list.size() - 1);        }    }}

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