LeetCode 40. Combination Sum II_一个树求组合目标值-优快云博客

本文链接：https://blog.youkuaiyun.com/nianhua120/article/details/52055123

本文介绍了一个经典的组合求和问题：给定一组候选数和目标数，寻找所有可能的组合使得候选数之和等于目标数。文章提供了一种递归算法，并通过在递归过程中去除重复项来提高效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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]
]

本来想着用Set来去除重，结果TLE啦。那只能在递归中的for循环里面去重了。

public class Solution {
    private void helper(int[] candidates, int pos, int target, Stack<Integer> stack, List<List<Integer>> list){
        if(target < 0) return ;
        if(target == 0){
            List<Integer> row = new ArrayList<>();
            for(int i:stack){
                row.add(i);
            }
            list.add(row);
            return;
        }
        int prev = 0;
        for(int i=pos; i<candidates.length; i++){
            if(candidates[i] == prev) continue;
            prev =candidates[i];
            stack.push(candidates[i]);
            helper(candidates, i+1, target-candidates[i], stack, list);
            stack.pop();
        }
        
    }
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        Stack<Integer> stack = new Stack<>();
        List<List<Integer>> lists = new ArrayList<>();
        Arrays.sort(candidates);
        helper(candidates, 0, target, stack, lists);

        return lists;
    }
}