leetcode 39. Combination Sum DFS深度优先搜索 + 按照index做递归

亦凉 2022-06-09 09:46 132阅读 0赞

Given a set of candidate numbers (C) (without duplicates) 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:  
\[  
\[7\],  
\[2, 2, 3\]  
\]

题意就是说在一个set中寻找元素，使和为target，**注意元素可以重复使用**。典型的DFS深度优先搜索去做。

建议和这一道题[leetcode 377. Combination Sum IV 组合之和 + DP动态规划 + DFS深度优先遍历][leetcode 377. Combination Sum IV _ _ DP_ _ DFS]、[leetcode 216. Combination Sum III DFS + 按照index递归遍历][leetcode 216. Combination Sum III DFS _ _index] 、[leetcode 40. Combination Sum II DFS深度优先搜索][leetcode 40. Combination Sum II DFS] 一起学习

**注意这个是包含当前元素的**

代码如下：

import java.util.ArrayList;
    import java.util.Arrays;
    import java.util.List;
    
    public class Solution 
    {
        
        public List<List<Integer>> combinationSum(int[] candidates, int target)
        {
            List<List<Integer>> res=new ArrayList<List<Integer>>();
            List<Integer> one=new ArrayList<>();
    
            //考虑特殊情况
            if(candidates==null || candidates.length<0)
                return res;
            else
            {
                Arrays.sort(candidates);
                getSum(candidates,target,res,one,0,0);
                return res;
            }
        }
    
        private void getSum(int[] candidates, int target, List<List<Integer>> res,
                List<Integer> one, int sum,int level) 
        {
            if(sum>target)
                return ;
            else if(sum==target)
            {
                res.add(new ArrayList<>(one));
            }else
            {
                //题目中说可以重复的使用某一个数，所以level可以不变
                for(int i=level;i<candidates.length;i++)
                {
                    one.add(candidates[i]);
                    //所以这里的level为i就是表示可以重复使用一个值
                    getSum(candidates, target, res, one, sum+candidates[i], i);
                    one.remove(one.size()-1);
                }
    
            }
    
        }
    }

下面的是C++的做法，就是一个经典的DFS深度优先搜索遍历的做法，很经典和典型，值得学习

其实上面的Java的递归做法逻辑上有点乱，一共两个临界情况：一个是Index一个是sum，那么直接分情况讨论即可

代码如下：

#include <iostream>
    #include <vector>
    #include <map>
    #include <unordered_map>
    #include <set>
    #include <unordered_set>
    #include <queue>
    #include <stack>
    #include <string>
    #include <climits>
    #include <algorithm>
    #include <sstream>
    #include <functional>
    #include <bitset>
    #include <numeric>
    #include <cmath>
    #include <regex>
    
    using namespace std;
    
    
    
    class Solution
    {
    public:
        vector<vector<int>> res;
        vector<vector<int>> combinationSum(vector<int>& can, int target)
        {
            sort(can.begin(), can.end());
            dfs(can, target, 0, 0, vector<int>());
            return res;
        }
    
        void dfs(vector<int>& can, int target, int sum, int index, vector<int> one)
        {
            if (index == can.size())
            {
                if (sum == target)
                    res.push_back(one);
                return;
            }
            else
            {
                if (sum > target)
                    return;
                else if (sum == target)
                    res.push_back(one);
                else
                {
                    for (int i = index; i < can.size(); i++)
                    {
                        one.push_back(can[i]);
                        dfs(can, target, sum + can[i], i, one);
                        one.pop_back();
                    }
                }
            }
        }
    };

[leetcode 377. Combination Sum IV _ _ DP_ _ DFS]: http://blog.csdn.net/jackzhang_123/article/details/78174174
[leetcode 216. Combination Sum III DFS _ _index]: http://blog.csdn.net/jackzhang_123/article/details/78059803
[leetcode 40. Combination Sum II DFS]: http://blog.csdn.net/jackzhang_123/article/details/77803003