leetcode -- 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, � , a_k) must be in non-descending order. (ie, a₁ ? a₂ ? � ? a_k).
• 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] 

本题与上题Combination Sum类似，只是添加了去重部分

input	output	expected
[1,1], 1	[[1],[1]]	[[1]]

public class Solution {
    public ArrayList<ArrayList<Integer>> combinationSum2(int[] num, int target) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
        int len = num.length, depth = 0;
        if(len == 0){
            return result;
        }
        ArrayList<Integer> output = new ArrayList<Integer>();
        int sum = 0;
        Arrays.sort(num);
        generate(result, output, sum, depth, len, target, num);
        return result;
    }
    
     public void generate(ArrayList<ArrayList<Integer>> result, ArrayList<Integer> output, int sum,
            int depth, int len, int target, int[] candidates){
            if(sum > target){
                return;
            }
            if(sum == target){
                ArrayList<Integer> tmp = new ArrayList<Integer>();
                tmp.addAll(output);
                result.add(tmp);
                return;
            }
            
            for(int i = depth; i < len; i++){
                sum += candidates[i];
                output.add(candidates[i]);
                generate(result, output, sum, i + 1, len, target, candidates);
                sum -= output.get(output.size() - 1);
                output.remove(output.size() - 1);
35                 while(i < len - 1 && candidates[i] == candidates[i+1])
36                     i++;
            }
        }
}