第 1 天

栈与队列（简单）

剑指 Offer 09. 用两个栈实现队列

class CQueue {
public:
    CQueue() {}
    
    void appendTail(int value) {
        s1.push(value);
    }
    
    int deleteHead() {
        while(!s1.empty())
        {
            s2.push(s1.top());
            s1.pop();
        }
        int res;
        if(!s2.empty())
        {
            res = s2.top();
            s2.pop();
        }
        else
            return -1;
        while(!s2.empty())
        {
            s1.push(s2.top());
            s2.pop();
        }
        return res;
    }
    stack s1;
    stack s2;
};

剑指 Offer 30. 包含min函数的栈

【解法一】一个栈存放一个结构体结点

typedef struct Node
{int num;int min;
}Node;
class MinStack {
public:/** initialize your data structure here. */MinStack() {}void push(int x) {Node a;a.num = x;a.min = x;if(!s.empty() && s.top().min s;
};

【解法二】使用俩个栈

class MinStack {
public:/** initialize your data structure here. */MinStack() {}void push(int x) {num.push(x);if(minnum.empty()){minnum.push(x); }else{if(minnum.top()>=x)minnum.push(x);}}void pop() {if(num.empty())return;else{if(num.top()==minnum.top())minnum.pop();num.pop();}}int top() {if(!num.empty())return num.top();return -1;}int min() {if(!num.empty())return minnum.top();return -1;}stack num;stack minnum;
};

第 2 天

链表（简单）

剑指 Offer 06. 从尾到头打印链表

【解法一】使用vector存储，并反转结果

class Solution {
public:vector reversePrint(ListNode* head) {ListNode* cur = head;vector res;while(cur){res.push_back(cur->val);cur = cur->next;}reverse(res.begin(), res.end());return res;}
};

【解法二】使用递归思想

class Solution {
public:vector res;vector reversePrint(ListNode* head) {if(!head)return res;reversePrint(head->next);res.push_back(head->val);return res;}
};

【解法三】使用一个辅助栈

class Solution {
public:vector reversePrint(ListNode* head) {vector res;if(!head)return res;stack s;while(head){s.push(head->val);head=head->next;}while(!s.empty()){res.push_back(s.top());s.pop();}return res;}
};

剑指 Offer 24. 反转链表

【解法一】迭代

class Solution {
public:ListNode* reverseList(ListNode* head) {if(!head)return head;ListNode* cur = head;ListNode* Next = head;ListNode* prev = NULL;while(cur){Next = cur->next;cur->next = prev;prev = cur;cur = Next;}return prev;}
};

【解法二】递归

class Solution {
public:ListNode* recur(ListNode* cur, ListNode* prev){if(cur==nullptr)return prev;ListNode* res = recur(cur->next, cur);cur->next = prev;return res;}ListNode* reverseList(ListNode* head) {return recur(head, nullptr);}
};

 剑指 Offer 35. 复杂链表的复制