剑指offer刷题记录

day1

剑指 Offer 09. 用两个栈实现队列 - 力扣（LeetCode）

class CQueue {
private:
    stack<int> PUSH;
    stack<int>POP;
public:
    void appendTail(int value) {
        PUSH.push(value);
    }
    
    int deleteHead() {
        if(POP.empty())
        {
            int x = 0;
            while(!PUSH.empty())
            {
                x = PUSH.top();
                POP.push(x);
                PUSH.pop();
            }
        }
        if(POP.empty())return -1;
        int x = POP.top();
        POP.pop();
        return x;
    }
};

剑指 Offer 30. 包含min函数的栈 - 力扣（LeetCode）

class MinStack {
private:
    stack<int>st;
    stack<int>_min;

public:
   void push(int x) {
       st.push(x);
       if(_min.empty() || _min.top()>=x)
        _min.push(x);
    }
    
    void pop() {
        int x = st.top();
        st.pop();
        if(x == _min.top())
        {
            _min.pop();
        }
    }    
    int top() {
        return st.top();
    }  
    int min() {
        return _min.top();
    }
};

day2

剑指 Offer 06. 从尾到头打印链表 - 力扣（LeetCode）

class Solution {
public:
    vector<int> reversePrint(ListNode* head) {
        ListNode *t = head;
        stack<int>st;
        vector<int>v;
        while(t != nullptr)
        {
            st.push(t->val);
            t = t->next;
        }

        while(!st.empty())
        {
            int x = st.top();
            v.push_back(x);
            st.pop();
        }
        return v;
    }
};

剑指 Offer 24. 反转链表 - 力扣（LeetCode）

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        stack<int>st;
        ListNode *t = head;
     
        while(t!=nullptr)
        {
            st.push(t->val);
            t = t->next;
        }
        
        ListNode *h = new ListNode;
        h ->next = NULL;
        h ->val = 0;
        t = h;
  
        while(!st.empty()) {
            int x = st.top();
            st.pop();
            
            ListNode *temp = new ListNode;
            temp->next = NULL;           
            temp->val = x;
            t->next = temp;
            t = temp;
        }
        return h->next;
    }
};

剑指 Offer 35. 复杂链表的复制 - 力扣（LeetCode）

class Solution {
public:
    Node* copyRandomList(Node* head) 
    {
        Node*t = head;
        
        unordered_map<Node*,Node*> m;
        while(t != nullptr)
        {
            m[t] = new Node(t->val);
            t = t->next;
        }

        t = head;
        while(t!=nullptr)
        {
            m[t] ->next = m[t->next];
            m[t] ->random = m[t->random];
            t = t->next;
        }
        return m[head];
    }  
};

day3

剑指 Offer 05. 替换空格 - 力扣（LeetCode）

class Solution {
public:
    string replaceSpace(string s) {
        string ans;
        for(auto x : s)
        {
            if(x == ' ')
                ans += "%20";
            else
                ans += x;
        }
        return ans;
    }  
};
//cout<<typeid(x).name()<<endl;得知x的类型是char

剑指 Offer 58 - II. 左旋转字符串 - 力扣（LeetCode）

class Solution {
public:
    string reverseLeftWords(string s, int n) {
        string s1 = s.substr(0,n);
        string s2 = s.substr(n);
        s  = s2 + s1;
        return s;
     }
};

day4

剑指 Offer 03. 数组中重复的数字 - 力扣（LeetCode）

class Solution {
public:
    int findRepeatNumber(vector<int>& nums) {
        int count[100006] = {0};
        for(auto x:nums)
        {
            ++count[x];
            if(count[x]>1)
            {
                return x;
            }
        }
        return -1;
    }
};

剑指 Offer 53 - I. 在排序数组中查找数字 I - 力扣（LeetCode）

class Solution {
public:
    int search(vector<int>& nums, int target) {
        int ans = 0;
        int pos = lower_bound(nums.begin(),nums.end(), target) - nums.begin();
        cout<<pos<<endl;
        while(pos != nums.size())
        {
            if(nums[pos] == target)
            {
                ++ans;
            }
            ++pos;
        }
        return ans;
    }
};
//注意这个函数返回的pos是下标，而不是迭代器

剑指 Offer 53 - II. 0～n-1中缺失的数字 - 力扣（LeetCode）

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int ans = 0;
        for(int i = 0; i <= nums.size(); ++i)
        {
            if(i != nums[i])
            {
                ans = i;
                break;
            }
        }
        return ans;
    }
};

day5

剑指 Offer 04. 二维数组中的查找 - 力扣（LeetCode）

class Solution {
public:
    bool findNumberIn2DArray(vector<vector<int>>& matrix, int target) {
        int i = 0;
        for(; i < matrix.size(); ++i)
        {
            auto it = lower_bound(matrix[i].begin(), matrix[i].end(), target);
            if(it != matrix[i].end()&&*it == target)//返回的it也可能是野指针，我们不能访问野指针
                break;
        }
        if(i < matrix.size())
            return true;
        return false;
    }   
};

剑指 Offer 11. 旋转数组的最小数字 - 力扣（LeetCode）

class Solution {
public:
    int minArray(vector<int>& numbers) {
        sort(numbers.begin(),numbers.end());
        return numbers[0];
    }
};

剑指 Offer 50. 第一个只出现一次的字符 - 力扣（LeetCode）

class Solution {
public:
    char firstUniqChar(string s) {
        unordered_map<char, int> kv;
        for(auto x:s)
        {
            kv[x]++;
        }
        for(auto x:s)
        {
            if(kv[x] == 1)
            {
                return x;
            }
        }
        return ' ';        
    }
};

day6

剑指 Offer 32 - I. 从上到下打印二叉树 - 力扣（LeetCode）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> levelOrder(TreeNode* root) {
        queue<TreeNode*>q;
        vector<int>ans;
        if(root)q.push(root);

        while(!q.empty())
        {
            TreeNode* t = q.front();
            if(t)
                ans.push_back(t->val);
            q.pop();
            if(t->left)q.push(t->left);
            if(t->right)q.push(t->right);
        }
        return ans;
    }
};

剑指 Offer 32 - II. 从上到下打印二叉树 II - 力扣（LeetCode）

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*>q;
        vector<vector<int>>ans;
      
        if(root)q.push(root);
        
        while(!q.empty())
        {
            int size = q.size();
            vector<int> temp;

            for(int i=0;i<size;++i)
            {
                TreeNode* t = q.front();    
                q.pop();            
                temp.push_back(t->val); 
                if(t->left)q.push(t->left);
                if(t->right)q.push(t->right);
            }
            ans.push_back(temp);
        }
        return ans;
    }
};

剑指 Offer 32 - III. 从上到下打印二叉树 III - 力扣（LeetCode）

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*>q;
        vector<vector<int>>ans;
        bool f = false;
        if(root)q.push(root);
        
        while(!q.empty())
        {
            int size = q.size();
            vector<int> temp;

            for(int i=0;i<size;++i)
            {
                TreeNode* t = q.front();    
                q.pop();            
                temp.push_back(t->val); 
                if(t->left)q.push(t->left);
                if(t->right)q.push(t->right);
            }
            if(f==true)reverse(temp.begin(),temp.end());
            ans.push_back(temp);
            f = !f;
        }
        return ans;
    }
};