代码随想录算法训练营第28天|93.复原IP地址 78.子集 90.子集II

93.复原IP地址

class Solution {
private:
    vector res;
    void backtracking(string& s, int startIndex, int pointSum) {
        if (pointSum == 3) {
            if (isvalid(s, startIndex, s.size() - 1)) {
                res.push_back(s);
            }
            return;
        }
        for (int i = startIndex; i < s.size(); i++) {
            if (isvalid(s, startIndex, i)) {
                s.insert(s.begin() + i + 1, '.');
                pointSum++;
                backtracking(s, i + 2, pointSum);
                s.erase(s.begin() + i + 1);
                pointSum--;
            }else break;
        }
    }
    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;
    }
public:
    vector restoreIpAddresses(string s) {
        res.clear();
        if (s.size() < 4 || s.size() > 12) return res;
        backtracking(s, 0, 0);
        return res;
    }
};

 for循环中递归参数要i + 2，跳过逗点。

注意.insert()以及.erase()函数的运用来添加删除逗点。

78.子集

class Solution {
private:
    vector path;
    vector> res;
    void backtracking(vector nums, int startIndex) {
        res.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> subsets(vector& nums) {
        path.clear();
        res.clear();
        backtracking(nums, 0);
        return res;
    }
};

90.子集II

class Solution {
private:
    vector> result;
    vector path;
    void backtracking(vector& nums, int startIndex, vector& used) {
        result.push_back(path);
        for (int i = startIndex; i < nums.size(); i++) {
            // used[i - 1] == true，说明同一树枝candidates[i - 1]使用过
            // used[i - 1] == false，说明同一树层candidates[i - 1]使用过
            // 而我们要对同一树层使用过的元素进行跳过
            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);
            used[i] = false;
            path.pop_back();
        }
    }

public:
    vector> subsetsWithDup(vector& nums) {
        result.clear();
        path.clear();
        vector used(nums.size(), false);
        sort(nums.begin(), nums.end()); // 去重需要排序
        backtracking(nums, 0, used);
        return result;
    }
};