LeetCode 93.复原IP地址、LeetCode 78.90.子集

93. 复原 IP 地址

方法：回溯

class Solution {
private:
    vector<string> res;
    void dfs(string& s, int idx, int pn) {
        if (pn == 3) {
            if (isVaild(s, idx, s.size()-1)) res.emplace_back(s);
            return ;
        }
        for (int i = idx; i < s.size(); ++i) {
            if (isVaild(s, idx, i)) {
                s.insert(s.begin()+i+1, '.');
               
                dfs(s, i+2, pn+1);
              
                s.erase(s.begin()+i+1);
            } else break;
        }
    }

    bool isVaild(const string& s, int st, int ed) {
        if (st > ed) return false;
        if (s[st] == '0' && st != ed) return false;
        int num = 0;
        for (int i = st; i <= ed; ++i) {
            if (s[i] < '0' || s[i] > '9') return false;
            num = num * 10 + (s[i] - '0');
            if (num > 255) return false;    
        }
        
        return true;
    }
public:
    vector<string> restoreIpAddresses(string s) {
        res.clear();
        if (s.size() < 4 || s.size() > 12) return res;
        dfs(s, 0, 0);
        return res;
    }
};

时间复杂度O(|s|)，空间复杂度O(1)

78. 子集

方法:回溯

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

时间复杂度O(n)，空间复杂度O(n)

90. 子集 II

方法:回溯

class Solution {
private:
    vector<vector<int>> res;
    vector<int> path;
    void dfs(vector<int>& nums, int idx) {
        res.emplace_back(path);
        if (idx >= nums.size()) return ;
        for (int i = idx; i < nums.size(); ++i) {
            if (i > idx && nums[i] == nums[i-1]) continue;
            path.emplace_back(nums[i]);
            dfs(nums, i + 1);
            path.pop_back();
        }
    }
public:
    vector<vector<int>> subsetsWithDup(vector<int>& nums) {
        res.clear();
        path.clear();
        sort(nums.begin(), nums.end());
        dfs(nums, 0);
        return res;
    }
};

时间复杂度O(n)，空间复杂度O(n)