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LeetCode 40. Combination Sum II

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Question

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
2
3
4
5
6
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Explanation

其实相比1还要简单一些。用DFS找到所有的组合，注意去重。

Code

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public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> list = new ArrayList<>();
        if (candidates == null || candidates.length == 0) return list;
        Arrays.sort(candidates);
        helper(candidates, 0, target, new ArrayList<Integer>(), list);
        return list;
    }
    // 在list[index:]中选取和为target的所有组合，添加到combination之后，整个放到combinations当中
    public void helper(int[] candidates, int startIndex, int target, List<Integer> combination, List<List<Integer>> list) {
        if (target == 0) list.add(new ArrayList<Integer>(combination));
        for(int i=startIndex; i< candidates.length; i++) {
            if (i >startIndex && candidates[i] == candidates[i-1]) continue; //去重的重要步骤，因为数列是升序。
            if (target < candidates[i]) {
                break;
            }
            combination.add(candidates[i]);
            helper(candidates, i+1, target - candidates[i], combination,list);
            combination.remove(combination.size()-1);
        }
    }
}