代码随想录算法训练营第二十四天|93 复原IP地址、78 子集、 90 子集II

 一、93 复原IP地址

class Solution {
private:
    vector<string>result;
    void backtracking(string &s,int startindex,int pointsum){
        if(pointsum == 3){
            //添加逗点的数量为3
            if(isValid(s,startindex,s.size()-1)){
                result.push_back(s);
                return;
            }
        }
        for(int i=startindex;i<s.size();i++){
            if(isValid(s,startindex,i)){
                s.insert(s.begin()+i+1,'.');
                pointsum += 1;
                backtracking(s,i+2,pointsum);
                s.erase(s.begin()+i+1);
                pointsum -= 1;
            }
        }
    }
    bool isValid(const string &s,int begin,int end){//判断点前的字符串是否合法
        if(begin>end)
            return false;
        if(s[begin]=='0'&&begin!=end)//0开头的
            return false;
        int num = 0;
        for(int i=begin;i<=end;i++){
            if(s[i]>'9'||s[i]<'0'){//非数字的
                return false;
            }
            num = num*10+(s[i]-'0');
            if(num>255){//如果大于255，不合法
                return false;
            }
        }
        return true;
    }
public:
    vector<string> restoreIpAddresses(string s) {
        if(s.size()<4||s.size()>12)//剪枝
            return result;
        backtracking(s,0,0);
        return result;
    }
};

二、78 子集

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

三、 90 子集II  

class Solution {
private:
    vector<vector<int>>result;
    vector<int>path;
    void backtracking(vector<int>&nums,int startindex,vector<bool>&used){
        result.push_back(path);
        if(startindex >= nums.size())
            return;
        for(int i = startindex;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);
            path.pop_back();
            used[i] = false;
        }
    }
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        vector<bool>used(nums.size(),false);
        sort(nums.begin(),nums.end());//去重一定要排序
        backtracking(nums,0,used);
        return result;
    }
};