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.
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]
]

跟上一篇思想其实是一样的，只不过这个不能用同一水平线上即不能重复一个元素。上代码，递归自己写出来，看来还是懂了，嘿嘿。

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (candidates.length == 0)
            return res;
        List<Integer> list = new ArrayList<Integer>();
        Arrays.sort(candidates);
        findSum(candidates, list, 0, 0, target, res);
        return res;
    }

    private void findSum(int[] candidates, List<Integer> list, int sum,
            int level, int target, List<List<Integer>> res) {
        if (sum == target) {
            if (!res.contains(list))
                res.add(new ArrayList<Integer>(list));
            return;
        } else if (sum > target)
            return;
        else
            for (int i = level; i < candidates.length; i++) {
                list.add(candidates[i]);
                findSum(candidates, list, sum + candidates[i], i+1, target, res);
                list.remove(list.size() - 1);
            }
    }
}