Leetcode 40 Combination Sum II

最新推荐文章于 2024-11-20 10:20:22 发布

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最新推荐文章于 2024-11-20 10:20:22 发布

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本文链接：https://blog.youkuaiyun.com/Neo233/article/details/83988494

本文探讨了一道经典的算法题目——组合求和II，旨在找到所有可能的候选数组合，使得组合中数字之和等于给定的目标值。文章提供了两种解决方案，一种是通过递归实现，另一种则利用深度优先搜索(DFS)算法进行求解，同时确保了组合的唯一性和不重复性。

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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 1:

Input: candidates = [10,1,2,7,6,1,5], target = 8,
A solution set is:
[
  [1, 7],
  [1, 2, 5],
  [2, 6],
  [1, 1, 6]
]

Example 2:

Input: candidates = [2,5,2,1,2], target = 5,
A solution set is:
[
  [1,2,2],
  [5]
]

这个题是上一个题的变形，使用list来去重并且将变量每次加一就可以了

1）

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

2）

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);
        List<Integer> path = new ArrayList<>();
        dfs(list,path,candidates,target,0);
        return list;
    }
    public void dfs(List<List<Integer>> list,List<Integer> path,int[] candidates,int target,int remain){
        if(target < 0){
            return ;
        }else if(target == 0){
            if(!list.contains(path)){
                list.add(new ArrayList<>(path));
            }
        }
        for(int i = remain ; i < candidates.length ; ++i){
            path.add(candidates[i]);
            dfs(list,path,candidates,target - candidates[i],i+1);
            path.remove(path.size() - 1);
        }
    }
}