代码随想录day24

93.复原IP地址，需二刷

//需理解回溯分割IP的做法以及IP段是否合法的实现

    vector<string> result;
    bool isValid(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] > '9' || s[i] < '0'){
                return false;
            }
            num = num * 10 + (s[i] - '0');
            if( num > 255){
                return false;
            }
        }
        return true;
    }
    void backtracking(string& s, int start, int point){
        if(point == 3){
            if(isValid(s, start, s.size()-1)){
                result.push_back(s);
            }
            return;
        }
        for(int i = start; i < s.size(); i++){
            if(isValid(s, start, i)){
                s.insert(s.begin() + i + 1 , '.');
                point++;
                backtracking(s, i+2, point);
                s.erase(s.begin() + i + 1);
                point--;
            }else break;
            
        }
    }
    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

//分割问题用组合回溯

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

90.子集II

    vector<vector<int>> result;
    vector<int> path;
    
    void backtracking(vector<int>& nums, int start, vector<bool>& used){
        result.push_back(path);
        if(start >= nums.size()){
            return;
        }
        for(int i = start; i < nums.size(); i++){
            if(i>0 && (nums[i] == nums[i-1]) && (used[i-1] == false)){
                continue;
            }
            path.push_back(nums[i]);
            used[i] = true;
            backtracking(nums, i+1, used);
            path.pop_back();
            used[i] = false;
        }
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        result.clear();
        path.clear();
        vector<bool> used(nums.size(), false);
        sort(nums.begin(), nums.end());
        backtracking(nums, 0, used);
        return result;
    }