代码随想录Day 28| 93.复原IP地址 | 78.子集 | 90.子集II

93. 复原 IP 地址 

class Solution {
public:
    vector<string> result;
    vector<string> restoreIpAddresses(string s) {
        backtracking(s,0,0);
        return result;
    }
    void backtracking(string&s, int startIndex, int pointNum){
        if(pointNum == 3){  //分割符到3，进行一次判断，如果判断通过，则加入结果集
            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,'.');
                pointNum++;
                backtracking(s,i+2,pointNum);
                pointNum--;
                s.erase(s.begin()+i+1);
            }
            else break;
        }
    }
    bool isValid(const string& s, int start, int end){
        if(start>end){
            return false;
        }
        if(s[start] =='0' && start!=end){  //不能含有前导0，否则返回false
            return false;
        }
        int num=0;
        for(int i = start;i<=end;i++){
            if(s[i]>'9'|| s[i]<'0'){    //判断是否含有非数字字符
                return false;
            }
            num = num *10 +(s[i]-'0');   //累加
            if(num>255){     //判断是否大于255，否则无效
                return false;
            }
        }
        return true;
    }
};

78. 子集

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

90. 子集 II

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    int used[11];
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        backtracking(nums,0);
        return res;
    }
    void backtracking(vector<int>&nums, int index){
        res.push_back(path);
        for(int i=index;i<nums.size();i++){
            if(i&& nums[i-1]==nums[i] && used[i-1]==0)
            continue;
            used[i]=1;
            path.push_back(nums[i]);
            backtracking(nums,i+1);
            path.pop_back();
            used[i]=0;
        }
    }
    
};

比较暴力的做法：先正常回溯，再对结果进行去重，代码如下：

class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;
    int used[11];
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        backtracking(nums,0);
        sort(res.begin(),res.end());
        res.erase(unique(res.begin(),res.end()),res.end());
        return res;
    }
    void backtracking(vector<int>&nums, int index){
        res.push_back(path);
        for(int i=index;i<nums.size();i++){
            if(used[i])
            continue;
            used[i]=1;
            path.push_back(nums[i]);
            backtracking(nums,i+1);
            path.pop_back();
            used[i]=0;
        }
    }
    
};