leetcode 40. Combination Sum II

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本文介绍了一种基于深度优先搜索(DFS)的算法，用于从候选数字集合中寻找所有可能的唯一组合，这些组合的数字之和等于目标值。特别强调了候选数字只能使用一次，且解决方案中不允许出现重复组合。

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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]
]

和之前一些题方法类似，DFS

public class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> ans = new ArrayList<>();
        Arrays.sort(candidates);
        helper(ans, new ArrayList(), candidates, 0, target);
        return ans;
    }
    
    private void helper(List<List<Integer>> ans, List<Integer> list, int[] candidates, int start, int target){
        if(target==0){
            ans.add(new ArrayList(list));
            return;
        }else if(target<0){
            return;
        }
        for(int i=start; i<candidates.length; i++){
            if(i>start && candidates[i]==candidates[i-1]) continue;
            list.add(candidates[i]);
            helper(ans, list, candidates, i+1, target-candidates[i]);
            list.remove(list.size()-1);
        }
    }
}