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]

与[url=http://kickcode.iteye.com/blog/2274838]Combination Sum[/url]不同的是，这里数组中的元素只能使用一次，所以我们只要在寻找的过程中开始点从当前点的下一个元素开始就可以了，其他保持不变。还要注意的一点是可能会有重复元素，产生两个一样的结果，我们只需要保存一个就可以了。代码如下：

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        LinkedList<Integer> list = new LinkedList<Integer>();
        LinkedList<List<Integer>> llist = new LinkedList<List<Integer>>();
        if(candidates == null || candidates.length == 0) return llist;
        java.util.Arrays.sort(candidates);
        getCombination(0, target, candidates, list, llist);
        return llist;
    }
    public void getCombination(int start, int target, int[] candidates, LinkedList<Integer> list,
    LinkedList<List<Integer>> llist) {
        for(int i = start; i < candidates.length; i++) {
            if(target > candidates[i]) {
                list.add(candidates[i]);
                getCombination(i + 1, target - candidates[i], candidates, list, llist);
                list.removeLast();
            } else if(target == candidates[i]) {
                list.add(candidates[i]);
                if(!llist.contains(list))
                    llist.add(new LinkedList<Integer>(list));
                list.removeLast();
            } else {
                return;
            }
        }
    }
}