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. 蕴含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. 简单链表的复制 - 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;得悉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. 数组中反复的数字 - 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; }};//留神这个函数返回的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，咱们不能拜访野指针 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. 第一个只呈现一次的字符 - 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; }};