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

https://leetcode.cn/problems/restore-ip-addresses/

和前面那个回文串差不多，就是找各种方法来切割，然后判断是否合法就行。

class Solution {
public:
    bool isValid(string s) {
        if(s.length() > 1 && s[0] == '0') return false;
        int x = stoi(s);
        if (x < 0 || x > 255) return false;
        return true;
    }
    vector result;
    vector path;
    void backTracking(string s) {
        if (s.empty() && path.size() == 4) {
            result.push_back(path[0] + '.' + path[1] + '.' + path[2] + '.' + path[3]);
            return;
        }
        if (s.empty()) return;
        if (path.size() == 4) return;
        for (int i = 0; i < (s.length() < 3 ?  s.length() : 3); i++) {
            string left(s.begin(), s.begin() + i + 1);
            if (!isValid(left)) continue;
            path.push_back(left);
            string right(s.begin() + i + 1, s.end());
            backTracking(right);
            path.pop_back();
        }
        
    }
    vector restoreIpAddresses(string s) {
        backTracking(s);
        return result;
    }
};

78.子集

https://leetcode.cn/problems/subsets/

这道题我一开始想的很复杂

思路是这样的，先限定一个path大小的值k，回溯得到大小为k的所有子集添加到result中，然后依次从k=0到k=nums.size()全部回溯一遍

class Solution {
public:
    vector> result;
    vector path;
    void backTracking(vector nums, int k) {
        //k : 还需要再path中添加的元素数量
        if (k == 0) {
            result.push_back(path);
            return;
        }
        if (nums.empty()) {
            return;
        }
        for (int i = 0; i < nums.size(); i++) {
            //i : 添加到path中的元素位置
            path.push_back(nums[i]);
            vector nums2(nums.begin() + i + 1, nums.end());
            backTracking(nums2, k - 1);
            path.pop_back();
        }

    }
    vector> subsets(vector& nums) {
        for (int i = 0; i <= nums.size(); i++) {
            backTracking(nums, i);
        }
        return result;
    }
};

 后面发现其实就正常的回溯，只不过收集的条件从path大小达到要求改为每一步都收集……

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

 直接操作数组：

class Solution {
public:
    vector> result;
    vector path;
    void backTracking(vector& nums) {
        result.push_back(path);
        if (nums.empty()) return;
        for(int i = 0; i < nums.size(); i++){
            path.push_back(nums[i]);
            vector nums2(nums.begin() + i + 1, nums.end());
            backTracking(nums2);
            path.pop_back();
        }
    }
    vector> subsets(vector& nums) {
        backTracking(nums);
        return result;
    }
};

90.子集II

https://leetcode.cn/problems/subsets-ii/

和上面的子集问题差不多，但是多了一个去重的需求

class Solution {
public:
    vector> result;
    vector path;
    void backTracking(vector nums, int cur) {
        result.push_back(path);
        if (cur >= nums.size()) return;
        for(int i = cur; i < nums.size(); i++) {
            if (i > cur && nums[i] == nums[i - 1]) continue;
            path.push_back(nums[i]);
            backTracking(nums, i + 1);
            path.pop_back();
        }
    }
    vector> subsetsWithDup(vector& nums) {
        sort(nums.begin(), nums.end());
        backTracking(nums, 0);
        return result;
    }
};

直接操作数组：

class Solution {
public:
    vector> result;
    vector path;
    void backTracking(vector nums) {
        result.push_back(path);
        if (nums.empty()) return;
        for(int i = 0; i < nums.size(); i++){
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            path.push_back(nums[i]);
            vector nums2(nums.begin() + i + 1, nums.end());
            backTracking(nums2);
            path.pop_back();
        }
    }
    vector> subsetsWithDup(vector& nums) {
        sort(nums.begin(), nums.end());
        backTracking(nums);
        return result;
    }
};

使用set去重：

class Solution {
public:
    vector> result;
    vector path;
    void backTracking(vector nums, int cur) {
        result.push_back(path);
        if (cur >= nums.size()) return;
        unordered_set set;
        for(int i = cur; i < nums.size(); i++) {
            if (!set.empty() && set.find(nums[i]) != set.end()) continue;
            set.insert(nums[i]);
            path.push_back(nums[i]);
            backTracking(nums, i + 1);
            path.pop_back();
        }
    }
    vector> subsetsWithDup(vector& nums) {
        vector used(nums.size(), false);
        sort(nums.begin(), nums.end());
        backTracking(nums, 0);
        return result;
    }
};

 使用used去重：

class Solution {
public:
    vector> result;
    vector path;
    vector used = {false};
    void backTracking(vector nums, int cur) {
        result.push_back(path);
        if (cur >= nums.size()) return;
        for(int i = cur; i < nums.size(); i++) {
            if (i >= 1 && nums[i] == nums[i-1] && used[i-1] == false) continue;
            path.push_back(nums[i]);
            used[i] = true;
            backTracking(nums, i + 1);
            path.pop_back();
            used[i] = false;
        }
    }
    vector> subsetsWithDup(vector& nums) {
        sort(nums.begin(), nums.end());
        backTracking(nums, 0);
        return result;
    }
};