LeetCode 93 recovery IP address 78 subset 90 subset-ii | Code Caprice 25th training camp day28

LeetCode 93 Restore IP address 2023.11.21

• Question link
• Code Caprice Explanation [Link]
  

class Solution {
    
    
public:
    //成员变量用于记录答案字符串
    vector<string> result;
    //判断字符串在左闭右闭区间[start, end]内是否合法
    bool isValid(string s,int start, int end)
    {
    
    
        //区间不合理时返回false
        if(start > end)
            return false;
        //0开头的数字不合法
        if(s[start] == '0' && start != end)
            return false;
        //值不合法或者大于255时返回false
        int num = 0;
        for(int i = start; i <= end; i++)
        {
    
    
            if(s[i] > '9' || s[i] < '0')
                return false;
            num = num*10 + (s[i] - '0');
            if(num > 255)
                return false;
        }
        return true;
    }
    //用了引用，start表示搜索位置，pointNum为已添加逗点数量
    void backtracking(string &s, int start, int pointNum)
    {
    
    
        //逗号数为3时，递归结束
        if(pointNum == 3)
        {
    
    
            //如果最后的字符串合法则加入到答案变量中返回
            if(isValid(s, start, s.size()-1))
                result.push_back(s);
            return;
        }
        //递归回溯遍历
        for (int i = start; i < s.size()-1; i++)
        {
    
    
            //判断区间[start, i]字符串如果合理则开始递归
            if(isValid(s, start, i))
            {
    
    
                //在合法字符串后加'.'，逗点数量也增加
                s.insert(s.begin()+i+1, '.');
                pointNum++;
                //递归,因为加了逗点，所以start= i+2
                backtracking(s, i+2, pointNum);
                //回溯，去除添加的逗点，并减少逗点数量
                s.erase(s.begin()+i+1);
                pointNum--;
            }
            else
                break;
        }
        
    }

    vector<string> restoreIpAddresses(string s) {
    
    
        result.clear();
        if(s.empty())
            return result;
        backtracking(s, 0, 0);
        return result;
    }
};

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LeetCode 78 subset 2023.11.21

• Question link
• Code Caprice Explanation [Link]
  

class Solution {
    
    
public:
    vector<int> cur;
    vector<vector<int>> result;
    void backtracking(vector<int> &nums, int start)
    {
    
    
        //将当前数组加入到答案中
        result.push_back(cur);
        //递归终止条件：全部遍历完了
        if(start == nums.size())
            return;
        //递归
        for (int i = start; i < nums.size(); i++)
        {
    
    
            cur.push_back(nums[i]);
            backtracking(nums, i+1);
            cur.pop_back();//回溯
        }
        
    }
    vector<vector<int>> subsets(vector<int>& nums) {
    
    
        cur.clear();
        result.clear();
        backtracking(nums, 0);
        return result;
    }
};

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LeetCode 90 subset-ii 2023.11.21

• Question link
• Code Caprice Explanation [Link]
  

class Solution {
    
    
public:
    //成员变量存储当前答案用vector，存储最终答案用set为了去重
    vector<int> cur;
    set<vector<int>> result;
    //递归回溯遍历
    void backtracking(vector<int> nums, int start)
    {
    
    
        //将当前子集存入最后答案中
        result.insert(cur);
        //如果遍历结束则返回
        if(start == nums.size())
            return;
        //遍历回溯
        for (int i = start; i < nums.size(); i++)
        {
    
    
            cur.push_back(nums[i]);
            backtracking(nums, i+1);
            cur.pop_back();
        }
        
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
    
    
        cur.clear();
        result.clear();
        //先排序，后递归回溯，避免伪重复答案，如1,4,4/4,1,4
        sort(nums.begin(), nums.end());
        backtracking(nums, 0);
        return vector<vector<int>>(result.begin(), result.end());
    }
};