LeetCode 39.Combination Sum，40. Combination Sum II，216. Combination Sum III

最新推荐文章于 2025-05-25 18:32:47 发布

原创最新推荐文章于 2025-05-25 18:32:47 发布 · 342 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#C++ #leetcode #Combination Sum

leetcode 专栏收录该内容

49 篇文章

订阅专栏

本文提供LeetCode上三道组合总和问题的解决方案，包括Combination Sum I、II及III，采用回溯法结合深度优先搜索解决不同约束下的组合问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

LeetCode 39.Combination Sum

题目链接：https://leetcode.com/problems/combination-sum/

题目描述：

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

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[
  [7],
  [2, 2, 3]
]

思路：一遇到这种用递归的就晕，不知从何处入手，不知怎么设计，应该直面困难吗！参考了dr的代码，自己写了一遍，在此做个笔记。。。

用回溯法，通常，我们将回溯法和DFS等同看待，可以用一个等式表示它们的关系：回溯法=DFS+剪枝。所以回溯法是DFS的延伸，其目的在于通过剪枝使得在深度优先搜索过程中如果满足了回溯条件不必找到叶子节点，就截断这一条路径，从而加速DFS。实际上，即使没有剪枝，DFS在从下层回退到上层的时候也是一个回溯的过程，通常这个时候某些变量的状态。DFS通常用递归的形式实现比较直观，也可以用非递归，但通常需要借组辅助的数据结构（比如栈）来存储搜索路径。

参考链接：http://blog.youkuaiyun.com/u012501459/article/details/46779021

http://www.tuicool.com/articles/2Y3iay3

代码：

class Solution {
public:
	vector<vector<int>> combinationSum(vector<int>& candidates, int target) {
		vector<vector<int>> res;
		vector<int> path;
		vector<vector<int>> result;
		int sum;int ind=0;
	    dfs(candidates,target,path,result,sum,ind);
	    return result;
	}
    void dfs(vector<int>& candidates,int target,vector<int> path,vector<vector<int>>& result,int sum,int ind)
    {
        for(int i=ind;i<candidates.size();i++)
        {
            sum+=candidates[i];
            if(sum<=target)
            {
                path.push_back(candidates[i]);
                if(sum==target)
                    result.push_back(path);
                if(sum<target)
                    dfs(candidates,target,path,result,sum,i);
                path.pop_back();
            }
            sum-=candidates[i];
        }
    }
};

40. Combination Sum II

题目链接：https://leetcode.com/problems/combination-sum-ii/

题目描述：

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]
]

代码：

class Solution {
public:
	vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {
		vector<vector<int>> re;
		vector<int> comb;
		sort(candidates.begin(),candidates.end());
		dfs(candidates, target, re, comb, 0, 0);
		return re;
	}
	void dfs(vector<int>& candidates, int target, vector<vector<int>>& re, vector<int> comb, int tmp, int start){
		for (int i = start; i<candidates.size(); i++){
		    if(i>start&&candidates[i]==candidates[i-1]) continue;
			if (tmp + candidates[i]<target){
				comb.push_back(candidates[i]);
				tmp += candidates[i];
				dfs(candidates, target, re, comb, tmp, i + 1);
				comb.pop_back();
			    tmp -= candidates[i];
			}
			else if(tmp + candidates[i]==target){
				comb.push_back(candidates[i]);
				re.push_back(comb);
				comb.pop_back();
				break;
			}
		}
	}
};

216. Combination Sum III

题目链接：https://leetcode.com/problems/combination-sum-iii/

题目描述：

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Example 1:

Input: k = 3, n = 7

Output:

[[1,2,4]]

Example 2:

Input: k = 3, n = 9

Output:

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

思路：用回溯法，DFS+剪枝，每一位都由1-9个数字组成，遍历所有组合

class Solution {
public:
    vector<vector<int>> combinationSum3(int k, int n) {
        vector<vector<int>> vec;
        vector<int> comb;
        bool flag=0;
        dfs(1,k,n,0,flag,comb,vec);
        return vec;
    }
    void dfs(int start,int k,int n,int tmp,bool& flag,vector<int> comb,vector<vector<int>> &vec){
        if(k==0){
            if(tmp==n){
                flag=1;
                vec.push_back(comb); 
            } 
            return;
        }
        if(tmp>n) return;
        for(int i=start;i<=9;i++){
            tmp+=i;
            comb.push_back(i);
            dfs(i+1,--k,n,tmp,flag,comb,vec);
            if(flag==1) break;
            tmp-=i;
            comb.pop_back();
        }
        flag=0;
    }
};