[刷题笔记28 回溯算法 part04]

回溯算法

● 93.复原IP地址
● 78.子集
● 90.子集II

组合总和

93.复原IP地址

法1：回溯

    vector<string> ans;

    bool isVaild(string& s,int start,int end){
        if (start > end) return false;
        if (s[start] == '0' && start != end) return false;
        int num = 0;
        //非0-255
        for (int i = start; i <= end; ++i) {
            //丢失 排除非法数字
            if (s[i] > '9' || s[i] < '0') return false;
            num = num * 10 + s[i] - '0';//此处是i位置的数字
            if (num > 255)
                return false;
        }
        return true;
    }
    void backTracking(string& s,int startIndex, int pointNum){
        if (pointNum == 3){
            if (isVaild(s,startIndex,s.size() - 1))
                ans.push_back(s);
        }
        //丢失 向vector<int>填充数据

        for (int i = startIndex; i < s.size(); ++i) {
            if (isVaild(s,startIndex,i)){
                s.insert(s.begin() + i + 1,'.');
                pointNum++;
                backTracking(s,i+2,pointNum);
                s.erase(s.begin() + i + 1);
                pointNum--;//回溯
            }else break;
        }
    }
    vector<string> restoreIpAddresses(string s) {
        ans.clear();
        //丢失剪枝
        if (s.size() < 4 || s.size() > 12)return ans;
        backTracking(s,0,0);
        return ans;
    }

子集

78.子集
法1： 回溯

  v vector<int> path;
    vector<vector<int>> ans;
    void backTracking(vector<int>& nums, int startIndex){
        ans.push_back(path);
//        if (startIndex >= nums.size())
        for (int i = startIndex; i < nums.size(); ++i) {
            path.push_back(nums[i]);
            backTracking(nums,i + 1);
            path.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        path.clear();
        ans.clear();
        backTracking(nums,0);
        return ans;
    }

子集II

90.子集II

法1：回溯

   vector<int> path;
    vector<vector<int>> ans;
    void backTracking(vector<int>& nums, int startIndex,vector<bool>& used){
       // if (std::find(ans.begin(), ans.end(),path))
        ans.push_back(path);
//        if (startIndex >= nums.size())
        for (int i = startIndex; i < nums.size(); ++i) {
            if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]){
                continue;
            }
            path.push_back(nums[i]);
            used[i] = true;
            backTracking(nums,i + 1,used);
            used[i] = false;
            path.pop_back();
        }
    }
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {

        path.clear();
        ans.clear();
        vector<bool> used(nums.size(), false);
        sort(nums.begin(), nums.end()); // 去重需要排序
        backTracking(nums,0,used);
        return ans;
    }