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 toT.

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] 

 

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
    	List<List<Integer>> res = new ArrayList<List<Integer>>();  
        List<Integer> item = new ArrayList<Integer>();  
        if (candidates == null || candidates.length==0) {
        	return res;
        }
        Arrays.sort(candidates);
        boolean[] visited = new boolean[candidates.length];
        solve(candidates,target, 0, item ,res, visited);
        return res;
    }

	private void solve(int[] candidates, int target, int start, List<Integer> item,
			List<List<Integer>> res, boolean[] visited) {
		// TODO Auto-generated method stub
		if (target < 0) {
			return;
		}
		if (target == 0) {
			res.add(new ArrayList<>(item));
			return;
		}
		for (int j = start; j < candidates.length; j++) {
			if (!visited[j]) {
				if (j > 0 && candidates[j] == candidates[j-1] && visited[j-1] == false) {
					continue;
				}
				item.add(candidates[j]);
				visited[j] = true;
				solve(candidates,target-candidates[j], j+1, item ,res, visited);
				visited[j] = false;
				item.remove(item.size()-1);
			}
		}
	}
}