/*
 Author:     King, wangjingui@outlook.com
 Date:       Dec 20, 2014
 Problem:    Combination Sum II
 Difficulty: Easy
 Source:     https://oj.leetcode.com/problems/combination-sum-ii/
 Notes:
 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] 

 Solution: ..Similar to Combination Sum I, except for line 42 && 44.
 */
public class Solution {
    public List<List<Integer>> combinationSum2(int[] num, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        Arrays.sort(num);
        ArrayList<Integer> path = new ArrayList<Integer>();
        combinationSumRe(num, target, 0, path, res);
        return res;
    }
    void combinationSumRe(int[] candidates, int target, int start, ArrayList<Integer> path, List<List<Integer>> res) {
        if (target == 0) {
            ArrayList<Integer> p = new ArrayList<Integer>(path);
            res.add(p);
            return;
        }
        for (int i = start; i < candidates.length && target >= candidates[i]; ++i) {
            if (i!=start && candidates[i-1] == candidates[i]) continue;
            path.add(candidates[i]);
            combinationSumRe(candidates, target-candidates[i], i+1, path, res);
            path.remove(path.size() - 1);
        }
    }
}