LeetCode 40. Combination Sum II

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博客围绕LeetCode题目展开，给定候选数字集合和目标数，要找出集合中数字和为目标数的所有唯一组合，且每个数字只能用一次。解题需排序处理有重复的数，还学习了大佬的物理去重方法。

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LeetCode 40. Combination Sum II

Description

Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target.

Each number in candidates 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.

Example

在这里插入图片描述

Code

java

class Solution {
    private List<List<Integer>> result;
    
    public void dfs(int[] candidates, int index, LinkedList list, int sum, int target) {
        if(sum == target) {
            result.add(new LinkedList<Integer>(list));
            return;
        }
        for(int i = index; i < candidates.length; i++) {
            if(i > index && candidates[i] == candidates[i-1]) continue;
            if(sum + candidates[i] <= target) {
                list.add(candidates[i]);
                dfs(candidates, i+1, list, sum+candidates[i], target);
                list.removeLast();
            }
        }
    }
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        LinkedList list = new LinkedList<Integer>();
        result = new LinkedList<List<Integer>>();
        Arrays.sort(candidates);
        dfs(candidates, 0, list, 0, target);
        return result;
    }
}