LeetCode -- Combination Sum II

题目描述：
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] 


就是从一个序列candidates中找出非重复元素的组合，要求和等于target。


思路：
本题与Combination Sum很类似，区别在于，回溯时，从Index处开始；而前者每次都从0开始。
1.本题解法依旧是回溯，使用一个数组arr在遍历nums时，添加nums[i]，进入递归，移除nums[i]。i∈[0,n)
2.使用index变量来track下次循环的起始位置。
3.使用哈希来去重


实现代码：

    public IList<IList<int>> CombinationSum2(int[] candidates, int target) 
    {
    	if(candidates == null || candidates.Length == 0){
		    return null;
	}


    	var arr = candidates.OrderBy(x=>x).ToList();
    	
    	var result = new Dictionary<string, IList<int>>();
    	Travel(arr ,new List<int>(), 0, target, result);


    	return new List<IList<int>>(result.Select(x=>x.Value));
    }




private void Travel(IList<int> candidates, IList<int> arr, int index, int target, Dictionary<string ,IList<int>> result){
	if(target == 0 ){
	    var key = string.Join(",", arr);
	    if(!result.ContainsKey(key)){
	        result.Add(key, arr.ToList());
	    }
		
		return ;
	}
	
	for(var i = index ;i < candidates.Count; i++){
		if(target < candidates[i]){
			return;
		}
		
		arr.Add(candidates[i]);
		Travel(candidates, arr, i + 1, target - candidates[i], result);
		arr.Remove(candidates[i]);
	}
}