[代码随想录]基本数据结构篇

1.数组篇

---

1.1 704-二分查找

704

class Solution {
public:
    int search(vector<int>& nums, int target)
    {
        int begin =0;
        int end = nums.size()-1;
        int mid;
        while(begin <= end)  // -1,0,3,5,9,12
        {
            mid = begin+(end - begin)/2;
            if(nums[mid] < target)
                begin = mid+1;
            else if(nums[mid] > target)
                end = mid-1;
            else
                return mid;          
           
        }
        return -1;
    }
};

---

1.2 27-移除数组

27

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int fast = 0,slow = 0;
        while(fast != nums.size())
        {
            if(nums[fast] == val)
            {
                fast++;
            }
            else
            {
                nums[slow] = nums[fast];
                fast++;
                slow++;
            }
        }
        return slow;
    }
};

---

1.3 977-有序数组的平方

977

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int left = 0;
        int right = nums.size()-1;
        int flag = right;
        vector<int> v(right+1);
        while(left <= right)
        {
            int leftq = nums[left]*nums[left] ;
            int rightq = nums[right]*nums[right];
            if(rightq >= leftq)
            {
                v[flag--] = rightq;
                right--;
            }
            else
            {
                v[flag--] = leftq;
                left++;
            }
        }
        return v;
    }
};

---

1.4* 209–长度最小的子数组(滑动窗口)

209

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int left = 0,right = 0,sum = 0,min = INT_MAX;
        while(right < nums.size())
        {
            sum+=nums[right]; //统计左右指针之间的数据和
            while(sum >= target) //当期间数据大于等于target时进入,左指针向右移动
            {
                int newmin = right - left + 1;
                min = newmin < min ? newmin : min; //期间数据个数
                sum -= nums[left];
                left++;
            }
            right++;
        }
        return min==INT_MAX ? 0 : min;
    }
};

时间复杂度为 O(N)

---

1.5* 59-螺旋矩阵II

59

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> vv(n,vector<int>(n,0));
        int num = 0; //计数
        int k = 0; //活跃数字(填数坐标)
        int i = 0; //k依附坐标
        int j = n - 1; //k依附坐标

        while(i <= j) //当为奇数时会k == i == j,循环进入却没有填入操作
        {
            k = i;
            while(k < j)
                vv[i][k++] = ++num;
            
            k = i;
            while(k < j)
                vv[k++][j] = ++num;

            k = j;
            while(k > i)
                vv[j][k--] = ++num;

            k = j;
            while(k > i)
                vv[k--][i] = ++num;

            ++i;
            --j;
        }
        if(n%2)
            vv[(n-1)/2][(n-1)/2] = ++num;

        return vv;
    }
};

---

2. 链表篇

2.1 203-移除链表元素

203

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:

    void Recursion(ListNode* prev,ListNode* cur,int val)
    {
        if(cur == nullptr)
            return;
        Recursion(cur,cur->next,val);
        if(cur->val == val)
        {
            ListNode* tmp = cur;
            prev->next = cur->next;
            cur = cur->next;
            delete tmp;
        }
    }
    ListNode* removeElements(ListNode* head, int val) {
        while(head != nullptr && head->val == val)
            head = head->next;
        if(head != nullptr && head->next != nullptr)
            Recursion(head,head->next,val);
        return head;
    }
};

---

2.2 707-设计链表

707

class MyLinkedList {
public:

    struct ListNode
    {
        int val;
        ListNode* next;
        ListNode(int val)
            :val(val)
            ,next(nullptr)
        {}
    };

    MyLinkedList() 
    {
        head = new ListNode(0);
        size = 0;
    }
    
    int get(int index) {
        if(size <= index) //判断是否可以进行循环
            return -1;

        ListNode* cur = head->next;
        while(index--)
            cur = cur->next;
        return cur->val;
    }
    
    void addAtHead(int val) {
        addAtIndex(0,val);
    }
    
    void addAtTail(int val) {
        addAtIndex(size,val);
    }
    
    void addAtIndex(int index, int val) {
        if(size < index) //判断是否可以进行循环
            return;
        ListNode* newnode = new ListNode(val);
        ListNode* cur = head;
        while(index--)
            cur = cur->next; //cur 在index前
        newnode->next = cur->next;
        cur->next = newnode;
        size++;
    }
    
    void deleteAtIndex(int index) {
        if(size <= index)
            return;

        ListNode* cur = head;
        while(index--)
            cur = cur->next;
        ListNode* del = cur->next;
        ListNode* delnext = del->next;
        cur->next = delnext;
        delete del;
        size--;
    }

private:
    ListNode* head; //头节点
    int size;
};

/**
 * Your MyLinkedList object will be instantiated and called as such:
 * MyLinkedList* obj = new MyLinkedList();
 * int param_1 = obj->get(index);
 * obj->addAtHead(val);
 * obj->addAtTail(val);
 * obj->addAtIndex(index,val);
 * obj->deleteAtIndex(index);
 */

---

2.3 206-反转链表

206

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    void Recursion(ListNode*&head,ListNode* prev,ListNode* cur)
    {
        if(!cur->next)
        {
            head = cur;
            head->next = prev;
            return;
        }
        Recursion(head,cur,cur->next);
        cur->next = prev;
    }
    ListNode* reverseList(ListNode* head) {
        if(!head || !head->next)
            return head;
        Recursion(head,nullptr,head);
        return head;
    }
};

---

2.4* 24-两两交换链表中的节点(跳针)

24

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    void Recursion(ListNode*& head,ListNode* prev,ListNode* cur,ListNode* next)
    {
        if(cur == nullptr || cur->next == nullptr) //当到头时,也就是cur在倒数第二个
            return;
        Recursion(head,cur->next,cur->next->next,next); //next一直停留,为第一组第二个
        next = cur->next; //next置为未交换前的后一组的第二个,隔指
        if(prev == nullptr) //执行初始最后一次递归
            head = next;
        else
            prev->next = next;
            
        cur->next = next->next;
        next->next = cur;
        return;
    }

    ListNode* swapPairs(ListNode* head) 
    {
        if(head == nullptr || head->next == nullptr) //如果只有<两个直接return
            return head;
        Recursion(head,nullptr,head,head->next); //prev当作空
        return head;
    }
};

---

2.5* 19-删除链表的倒数第N个节点(快慢指针)

19

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    //void Reucursion(ListNode* prev,ListNode* cur)
    //{
    //    
    //} 1 2 3 4 5 6 7 8
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* fast = head;
        ListNode* slow = head;
        while(n--) 
            fast = fast->next; //fast先走,然后再走
        if(!fast) //fast如果走到头了,倒过来就说明等同于头删
        {
            head = head->next;
            delete slow;
            return head;
        }

        while(fast->next) //fast第二次行动,直到走到尾
        {
            fast = fast->next;
            slow = slow->next;
        }
        //此时slow位置就是要删除节点的上一位
        //开始删除
        ListNode* tmp = slow->next;
        slow->next = tmp->next;
        delete tmp;
        
        return head;
    }
};

---

2.6* 链表相交

链表相交

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(!headA || !headB)
            return nullptr;
        ListNode* pa = headA,* pb = headB;
        while(pa!=pb)
        {
            pa = pa==nullptr? headB : pa->next;
            pb = pb==nullptr? headA : pb->next;
        }
        return pa;
    }
};

---

2.7 142-环形链表II

142

方法一:哈希表

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        unordered_set<ListNode*> circle;
        while(head)
        {
            if(circle.count(head))
                return head;
            circle.insert(head);
            head = head->next;
        }
        return nullptr;
    }
};

时间复杂度：O(N)，其中 N 为链表中节点的数目。我们恰好需要访问链表中的每一个节点。
空间复杂度：O(N)，其中 N 为链表中节点的数目。我们需要将链表中的每个节点都保存在哈希表当中。

方法二:快慢指针

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        ListNode *slow = head, *fast = head;
        while (fast) 
        {
            slow = slow->next;

            if(!fast->next)
                return nullptr;
            fast = fast->next->next;
            if(fast == slow)
            {
                ListNode* cmcross = head;
                while(cmcross != slow)
                {
                    slow = slow->next;
                    cmcross = cmcross->next;
                }
                return cmcross;
            }
        }
        return nullptr;
    }
};

时间复杂度：O(N)
空间复杂度：O(1)

fast走两步,slow走一步,相遇后:头节点和slow一起向前走,相遇即为循环节点

---

3. 哈希篇

3.1 242-有效的字母异位词

242

class Solution {
public:
    bool isAnagram(string s, string t) {
        int record[26] = {0};
        for(auto& e : s)
            record[e-'a']++;
        for(auto& e : t)
            record[e-'a']--;
        for(auto& e : record)
        {
            if(e > 0 || e < 0)
                return false;
        }
        return true;
    }
};

时间复杂度：O(n)，其中 n 为 s的长度。
空间复杂度：O(S)，其中 S 为字符集大小，此处 S=26

---

3.2* 1002-查找共用字符

1002

class Solution {
public:
    vector<string> commonChars(vector<string>& words) {
        int record[26] = {0};
        int size = words.size(); // 判断几个字符串
        vector<string> vs;

        for (auto e : words[0])
            record[e - 'a']++;

        for (int i = 1; i < size; ++i) // 修改为从1遍历到size-1
        {
            // 用临时数组存储当前单词中字符的出现次数
            int temp[26] = {0}; 
            for (auto e : words[i])
                temp[e - 'a']++;
                
            // 更新record数组，保留每个字符在所有单词中的最小出现次数
            for (int j = 0; j < 26; ++j)
                record[j] = min(record[j], temp[j]);
        }

        for (int i = 0; i < 26; ++i)
        {
            while (record[i] > 0)
            {
                // 将字符插入到vs中，出现次数由record[i]控制
                vs.push_back(string(1, 'a' + i));
                record[i]--;
            }
        }

        return vs;
    }
};

---

3.3 349-两个数组的交集

349

class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        unordered_set<int> start(nums1.begin(),nums1.end());
        unordered_set<int> result; //去重
        for(int num : nums2)
        {
            if(start.find(num) != start.end())
                result.insert(num);
        }
        return vector<int>(result.begin(),result.end());
    }
};

当数组数据确认量少时

class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        unordered_set<int> result_set; // 存放结果，之所以用set是为了给结果集去重
        int hash[1005] = {0}; // 默认数值为0
        for (int num : nums1) // nums1中出现的字母在hash数组中做记录
            hash[num] = 1;
            
        for (int num : nums2)  // nums2中出现话，result记录
            if (hash[num] == 1) 
                result_set.insert(num);
           
        return vector<int>(result_set.begin(), result_set.end());
    }
};

---

3.4 202-快乐数

202

class Solution {
public:
    int get(int num)
    {
        int sum = 0;
        while (num)
        {
            sum += (num % 10) * (num % 10);
            num /= 10;
        }
        return sum;
    }
    bool isHappy(int n) {
        unordered_set<int> us;
        int tmp = n; // 使用tmp保存当前的数值
        while (tmp != 1 && us.find(tmp) == us.end()) // 当tmp变为1或出现循环时停止
        {
            us.insert(tmp);
            tmp = get(tmp);
        }
        return tmp == 1; // 如果tmp最终等于1，说明是快乐数，返回true，否则返回false
    }
};

---

3.5 1-两数之和

1

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int,int> um; //key为值,value为坐标
        for(int i = 0; i < nums.size(); ++i)
        {
            auto it = um.find(target - nums[i]);
            if(it != um.end())
                return {i,it->second};

            um.insert(make_pair(nums[i],i));
        }
        return {};
    }
};

T:O(N) S:O(N)

---

3.6* 454-四数相加II

454

class Solution {
public:
    int fourSumCount(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3, vector<int>& nums4) {
        unordered_map<int,int> um; //key为值,value为出现次数
        for(int num1 : nums1)
            for(int num2 : nums2)
                um[num1+num2]++;

        int count = 0;
        for(int num3 : nums3)
            for(int num4 : nums4)
                if(um.find(0-(num3+num4)) != um.end())
                    count+=um[(0-(num3+num4))];
        return count;
    }
};

---

3.7 383-赎金信

383

class Solution {
public:
    bool canConstruct(string ransomNote, string magazine) {
        int hash[26] = {0};
        for(char e : magazine)
            hash[e - 'a']++;
        for(char e : ransomNote)
            if(--hash[e-'a'] < 0)
                return false;
        return true;
    }
};

---

4. 双指针篇

4.1 15-三数之和(双指针)

15

哈希其实过于复杂,实现起来不如双指针

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int size = nums.size();
        sort(nums.begin(),nums.end());
        vector<vector<int>> vv;
        for(int i = 0; i < size - 2; ++i) //i为固定的数据
        {
            //为避免重复答案,对固定数据进行判断
            if(i > 0 && nums[i] == nums[i-1])
                continue; 
            
            int left = i+1; //中间数据
            int right = size-1; //末尾数据

            while(left < right)
            {
                int sum = nums[i] + nums[left] + nums[right];
                if(sum > 0)
                    right--;
                else if(sum < 0)
                    left++;
                else
                {
                    vv.push_back({nums[i] , nums[left] , nums[right]});

                    //为了避免重复答案,要对下一组数据进行判断
                    while(left<right && nums[left] == nums[left+1])
                        left++;
                    while(left<right && nums[right] == nums[right-1])
                        right--;
                    left++;
                    right--;
                }
            }
        }
        return vv;
    }
};

---

4.2 18-四数之和

18

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> ans;
        int size = nums.size();
        for(int i = 0;i<size;++i) //固定第一个
        {
            if(i>0&&nums[i-1] == nums[i])
                continue;
            for(int j = i+1;j<size;++j) //固定第二个
            {
                if(j>i+1&&nums[j-1] == nums[j]) //j>i+1,防止 2 2 2 2 情况
                    continue;
                int left = j+1;
                int right = size-1;
                while(left<right)
                {
                    long long sum =(long long)nums[i]+nums[j]+nums[left]+nums[right];
                    if(sum < target)
                        left++;
                    else if(sum > target)
                        right--;
                    else
                    {
                        ans.push_back({nums[i],nums[j],nums[left],nums[right]});
                        while(left<right && nums[left] == nums[left+1])
                            left++;
                        while(left<right && nums[right] == nums[right-1])
                            right--;
                        right--;
                        left++;
                    }
                }
            }
        }
        return ans;
    }
};

---

5. 字符串篇

5.1 344-反转字符串

344

reverse函数也可以

class Solution {
public:
    void reverseString(vector<char>& s) {
        int size = s.size();
        for(int left = 0,right = size-1 ; left < right ; ++left,--right)
            swap(s[left],s[right]);
    }
};

---

5.2 541-反转字符串II

541

class Solution {
public:
    string reverseStr(string s, int k) {
        for (int i = 0; i < s.size(); i += (2 * k)) 
        {
            if (i + k <= s.size()) 
                reverse(s.begin() + i, s.begin() + i + k );
            else
                reverse(s.begin() + i, s.end());
        }
        return s;
    }
};

---

5.3* 151-反转字符串中的单词

151

class Solution {
public:
    string reverseWords(string s) {
        while(s[0] == ' ')
            s.erase(s.begin()); //去除开头多余空格
        reverse(s.begin(),s.end());//blue is sky the
        while(s[0] == ' ')
            s.erase(s.begin()); //去除开头多余空格

        int count = 0, size = 0, flag = 0;
        for(int i =0 ; i < s.size();++i)
        {
            count++; //表示当前为第几个字符
            size++; //表示这个单词有几个字符
            if(s[i] == ' ')
            {
                reverse(s.begin()+flag, s.begin()+flag+size-1);
                flag = count; //记录下一个单词的起始位置
                size = 0;
                while(s[i+1] == ' ') //删除中间多余空格
                    s.erase(s.begin()+i+1);
            }
        }
        reverse(s.begin()+flag,s.end());
        return s;
    }
};

---

5.4 182-动态口令

182

class Solution {
public: //1 2 3 4 5 6 7 -> 4 5 6 7 1 2 3
        //7 6 5 4 3 2 1 -> 7 6 5 4 1 2 3 -> 4 5 6 7 1 2 3
    string dynamicPassword(string password, int target) {
        int size = password.size();
        int newstart = size - target;
        reverse(password.begin(),password.end());
        reverse(password.begin(),password.begin()+newstart);
        reverse(password.begin()+newstart,password.end());
        return password;
    }
};

---

5.5* 459-重复的子字符串

459

class Solution {
public:
    bool repeatedSubstringPattern(string s) 
    {
        string tmp(s.begin(),s.end());
        tmp+=s;
        tmp.erase(tmp.begin());
        tmp.erase(tmp.end()-1);
        //删去头尾
        if(tmp.find(s) == string::npos)
            return false;
        //在tmp内寻找s,找到了就说明是重复组成的
        //abab ->(a)b abab a(b)
        //abcdabcd -> (a)bcd abcdabcd abc(d)
        return true;
    }
};

---

5.6* 28-找出字符串中第一个匹配项的下标(KMP)

28

class Solution {
public: 
    //aabaabaafa  aabaaf
    //前缀表的作用就是再两个字符串匹配失败时,能回到最近的匹配位置,记录能回到位置下标
    void getNext(int* next, const string& s) 
    {
        //-1 0  -1 0  1  -1
        //a  a  b  a  a  f
        int j = -1;
        next[0] = j;
        for(int i = 1; i < s.size(); i++) 
        { 
             //能进入就说明前面已经有重复的了
            //进不去说明和前面一样继续重复
            while (j >= 0 && s[i] != s[j + 1])  // 前后缀不相同了
                j = next[j]; // 向前回退

            if (s[i] == s[j + 1])  // 找到相同的前后缀
                j++;

            next[i] = j; // 将j（前缀的长度）赋给next[i]
        }
    }
    int strStr(string haystack, string needle) 
    {
        int size_ndl = needle.size();
        int size_hstk = haystack.size();

        int next[size_ndl];
        getNext(next, needle);
        int j = -1; // // 因为next数组里记录的起始位置为-1
        for (int i = 0; i < size_hstk; i++) 
        { 
            // 注意i就从0开始
            //当数据和前缀表有重合时且当下不对应
            //while放在if(匹配)前面,防止j一直++匹配成功后又进入循环判断下一个不匹配而重置j
            while(j >= 0 && haystack[i] != needle[j + 1])
                j = next[j]; // j 寻找之前匹配的位置 //重新加载前缀表

            if (haystack[i] == needle[j + 1])  // 匹配，j和i同时向后移动
                j++; // i的增加在for循环里
            
            if (j == (size_ndl - 1) )  // //当数据和前缀表完全重合
                return (i - size_ndl + 1);
        }
        return -1;
    }
};

---

6. 栈和队列篇

6.1 232-用栈实现队列

232

class MyQueue {
public:
    MyQueue() {

    }
    
    void push(int x) {
        s1.push(x);
    }
    
    int pop() {
        if(s2.empty()) //s2就相当于队列,s1存储数据,反转数据存入s2;
        {
            while(!s1.empty())
            {
                s2.push(s1.top());
                s1.pop();
            }
        }
        int ans = s2.top();
        s2.pop();
        return ans;
    }
    
    int peek() {
        int tmp = this->pop();
        s2.push(tmp);
        return tmp;
    }
    
    bool empty() {
        if(s1.empty() && s2.empty())
            return true;
        return false;
    }

private:
    stack<int> s1;
    stack<int> s2;
};

---

6.2 225-用队列实现栈

225

class MyStack {
public:
    MyStack() {

    }
    
    void push(int x) {
        if(!q1.empty())
            q1.push(x);
        else
            q2.push(x);
    }
    
    int pop() { //将其中一个队列中size-1个元素全部移入另一个队列,留下的元素pop
        int ans;
        if(!q1.empty()) //q1不为空,将元素移动到q2
        {
            while(q1.size() != 1)
            {
                q2.push(q1.front());
                q1.pop();
            }
            ans = q1.front();
            q1.pop();
        }
        else
        {
            while(q2.size() != 1)
            {
                q1.push(q2.front());
                q2.pop();
            }
            ans = q2.front();
            q2.pop();
        }
        return ans;
    }
    
    int top() {
        int tmp = this->pop();
        this->push(tmp);
        return tmp;
    }
    
    bool empty() {
        if(q1.empty() && q2.empty())
            return true;
        return false;
    }

private:
    queue<int> q1;
    queue<int> q2;
};

---

6.3 20-有效的括号

20

class Solution {
public://{([])}
    bool isValid(string s) {
        stack<char> st;
        for(char e : s)
        {
            if(e=='[' || e=='(' || e=='{') //左符号进入栈中
                st.push(e);
            else
            {
                if(st.empty())
                    return false;
                else if(e == ')' && st.top()=='(' 
                        || e == '}' && st.top() == '{' 
                        || e == ']' &&  st.top() == '[')
                    st.pop(); //找到对应的符号后删除左符号
                else
                    return false; //如果不符合这个规则,那么就不构成
            }
        }
        return st.empty(); //如果最后为空,则说明全部对应上了,否则即false
    }
};

---

6.4 1047-删除字符串中的所有相邻重复项

1047

class Solution {
public:
    string removeDuplicates(string s) {
        stack<int> st;
        for(char e : s)
        {
            if(st.empty() || e!=st.top())
                st.push(e);
            else
                st.pop();
        }
        //将栈中的正确答案转移至string类
        string ans;
        while(!st.empty())
        {
            ans+=st.top();
            st.pop();
        }
        //由于从栈中获取数据,因此答案要反转回来
        reverse(ans.begin(),ans.end());
        return ans;
    }
};

---

6.5 150-逆波兰表达式求值

150

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> st;
        for(string& e : tokens)
        {
            if(e=="+" || e=="*" || e=="/" || e=="-")//如果为符号,则取栈顶两个数字进行运算
            {
                int second = st.top();
                st.pop();
                int first = st.top();
                st.pop();
                //取栈顶两个数据
                if(e=="+")
                    st.push(first+second);
                else if(e=="-")
                    st.push(first-second);
                else if(e=="*")
                    st.push(first*second);
                else if(e=="/")
                    st.push(first/second);
            }
            else //如果为数字则入栈
                st.push(stoi(e));
        }
        return st.top();
    }
};

---

6.6* 239-滑动窗口最大值(双端队列)

239

class Solution {
public:
    void push(int num)
    {
        while(!dq.empty() && dq.back() < num) //dq中第一个元素始终保证为最大值
            dq.pop_back(); 
        dq.push_back(num);
    }

    void pop(int num)
    {
        if(!dq.empty() && dq.front() == num)
            dq.pop_front();
    }

    vector<int> maxSlidingWindow(vector<int>& nums, int k) {
        vector<int> ans;
        //创建第一组滑动窗口
        for(int i = 0; i<k; ++i)
            push(nums[i]);
        ans.push_back(dq.front()); //放入第一组滑动窗口的最大值

        //开始处理后面的窗口
        int tmp = k;
        while(tmp < nums.size())
        {
            pop(nums[tmp-k]); //尝试删除dq中最大数字(滑动窗口滑过了这个数字,需要更新,因此删除)
            push(nums[tmp++]); //尝试更新
            ans.push_back(dq.front());
        }
        return ans;
    }

private:
    deque<int> dq; //双端队列
};

---

6.7* 347-前K个高频元素(优先级队列)

347

class Solution {
public:
    struct SecondCmp
    {
        bool operator()(const pair<int,int>& x,const pair<int,int>& y)
        {
            return x.second<y.second;
        }
    };

    vector<int> topKFrequent(vector<int>& nums, int k) {
        vector<int> ans;
        unordered_map<int,int> um; //key:元素 value:出现次数
        for(auto& e : nums)
            um[e]++;

        //使优先级队列升序排列,c++20适用
        //priority_queue,vector>,decltype([]
        //(const pair& x,const pair& y){
        //    return x.second
        //})> pq;
        priority_queue<pair<int,int>,vector<pair<int,int>>,SecondCmp> pq;


        for(auto& e : um)
            pq.push(make_pair(e.first,e.second));
        
        while(k--)
        {
            ans.push_back(pq.top().first);
            pq.pop();
        }
        return ans;
    }
};

---