代码随想录算法训练营Day28|93.复原IP地址、78.子集、90.子集II

93.复原IP地址

题目链接：93.复原IP地址
文档链接：93.复原IP地址
视频链接：回溯算法如何分割字符串并判断是合法IP？| LeetCode：93.复原IP地址

C++实现

class Solution {
private:
    vector<string> result;
    bool isValid(const string& s, int start, int end){
        if(start > end){
            return false;            
        }
        if(s[start] == '0' && start != end){
            return false;
        }
        int num = 0;
        for(int i = start; i<=end;i++){
            if(s[i] < '0' || s[i] > '9'){
                return false;
            }
            num = num * 10 + (s[i]-'0');
            if(num > 255){
                return false;
            }
        }
        return true;
    }

    void backtracking(string& s, int startIdx, int pointNum){
        if(pointNum == 3){
            if(isValid(s, startIdx, s.size()-1)){
                result.push_back(s);                
            }
            return;
        }
        for(int i = startIdx; i<s.size();i++){
            if(isValid(s,startIdx, i)){
                s.insert(s.begin()+i+1,'.');
                pointNum++;
                backtracking(s, i+2, pointNum);
                pointNum--;
                s.erase(s.begin()+i+1);
            }
            else break;
        }
        return;
    }
public:
    vector<string> restoreIpAddresses(string s) {
        result.clear();
        if(s.size()<4||s.size()>12) return result;
        backtracking(s, 0, 0);
        return result;
    }
};

78.子集

题目链接：78.子集
文档链接：78.子集
视频链接：回溯算法解决子集问题，树上节点都是目标集和！ | LeetCode：78.子集

C++实现

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, int startIdx){
        result.push_back(path);
        if(startIdx >= nums.size()) return;
        for(int i = startIdx; i<nums.size(); i++){
            path.push_back(nums[i]);
            backtracking(nums, i+1);
            path.pop_back();
        }
        return;
    }
public:
    vector<vector<int>> subsets(vector<int>& nums) {
        result.clear();
        path.clear();
        backtracking(nums, 0);
        return result;
    }
};

90.子集II

题目链接：90.子集II
文档链接：90.子集II
视频链接：回溯算法解决子集问题，如何去重？| LeetCode：90.子集II

C++实现

class Solution {
private:
    vector<vector<int>> result;
    vector<int> path;
    void backtracking(vector<int>& nums, int startIdx, vector<bool> used){
        result.push_back(path);
        if(startIdx>=nums.size()) return;
        for(int i = startIdx; i<nums.size(); i++){
            if(i>0 && nums[i]==nums[i-1] && used[i-1] == 0) continue;
            path.push_back(nums[i]);
            used[i] = true;
            backtracking(nums, i+1, used);
            used[i] = false;
            path.pop_back();
        }
        return;
    }

public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        result.clear();
        path.clear();
        sort(nums.begin(), nums.end());
        vector<bool> used(nums.size(), false);
        backtracking(nums, 0, used);
        return result;
    }
};