leetcode 206|92. Reverse Linked List 1|2-优快云博客

本文介绍两种链表反转方法：整体反转和区间[m, n]内的元素反转。通过具体实例展示了如何利用栈和直接指针操作实现链表的高效反转。

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206. Reverse Linked List

Reverse a singly linked list.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) 
    {
        //way-1
        /*
        ListNode *p=head;
        stack<int> cun;
        while(p)
        {  
            cun.push(p->val);
            p=p->next;
        }
        p=head;
        while(p)
        {  
            p->val=cun.top();
            cun.pop();
            p=p->next;
        }
        return head;
        */
        
        //way-2
        if(!head || !head->next)
            return head;
            
        ListNode *last=head;
        
        ListNode *p=last->next;
        ListNode *pre=p->next;
        
        while(p)
        {
            last->next = pre;
            p->next = head;
            head = p;
            
            p = last->next;
            if (p)
                pre = p->next;          
        }
        return head;
    }
};

92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

way-1:vector save

way-2:directly reverse list

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) 
    {
        //way-1
        /*
        ListNode *p=head;
        for(int i=1;i<m;i++)
            p=p->next;
        ListNode *q=p;
        stack<int> cun;
        for(int j=0;j<=n-m;j++)
        {
            cun.push(q->val);
            q=q->next;
        }
        q=p;
        for(int j=0;j<=n-m;j++)
        {
            q->val=cun.top();
            q=q->next;
            cun.pop();
        }
        return head;
        */
        
        //way-2
        if(!head || !head->next || m>=n)
            return head;
        
        ListNode hhead(-1);
        hhead.next=head;
        
        ListNode *point=&hhead;
        
        for(int i=1;i<m;i++)
            point=point->next;
        
        ListNode *p=point->next;
        ListNode *q=p->next;
        
        
        int count=n-m;
        while(count>0)
        {
            p->next=q->next;
            q->next=point->next;
            point->next=q;
            q=p->next;
            
            cout<<count--<<endl;
        }
        
        return hhead.next;
    }
};