40 Combination Sum II

/*
 * 先对candidates中的元素按从小到大排序，能节省时间
 */
public class Solution {
  List<List<Integer>> res = new ArrayList<List<Integer>>();
	public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        	if(candidates==null||candidates.length==0) return res;
        	int len = candidates.length;
        	int[] visited = new int[len];
        	Arrays.fill(visited, 0);
        	Arrays.sort(candidates);
        	dfs(new ArrayList<Integer>(), candidates, 0, target, visited);
        	return res;
    }
	public void dfs(List<Integer> list, int[] candidates, int sum, int target, int visited[]){
			int len = candidates.length;
			for(int i=0;i<len;++i){
				if(visited[i]==1) continue;
				if(sum + candidates[i]==target){
					List<Integer> tmp = new ArrayList<Integer>(list);
					tmp.add(candidates[i]);
					Collections.sort(tmp);
					if(!res.contains(tmp)){
						res.add(tmp);
					}
				}else if(sum + candidates[i] < target){
					List<Integer> tmp = new ArrayList<Integer>(list);
					tmp.add(candidates[i]);
					int[] tv = visited.clone();
					tv[i] = 1;
					dfs(tmp, candidates, sum + candidates[i], target, tv);
				}else{
				    /*
					 * 因为已经把candidates数组按从小到大排序，所以当sum + candidates[i] > target
					 *后，candidates的i以后的元素都不用试了，肯定不满足
					 */
					break;
				}
			}
	}
}