leetcode 92. Reverse Linked List II 反转链表 + 头插入反转链表（或者使用栈）

梦里梦外; 2022-06-08 02:38 114阅读 0赞

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.

这道题很简单，但是需要细心。

对于链表等问题都需要细心。

代码如下：

/*class ListNode
    {
          int val;
          ListNode next;
          ListNode(int x) { val = x; }
    }
    */
    /*
     * 链表反序问题，这个很简单，但是需要很细心
     * */
    public class Solution
    {
        public ListNode reverseBetween(ListNode head, int m, int n)
        {
            if(head==null || head.next==null)
                return head;
    
            ListNode newHead=new ListNode(0);
            newHead.next=head;
            ListNode pre=newHead,start=null,then=null;
            for(int i=0;i<m-1;i++)
                pre=pre.next;
    
            start=pre.next;
            then=start.next;
            for(int i=0;i<n-m;i++)
            {
                start.next=then.next;
                then.next=pre.next;
                pre.next=then;
                then=start.next;
            }
            return newHead.next;
    
        }    
    }

下面是C++的做法，就是一个简单的链表反转问题，可以直接使用头部插入法反转，或者使用栈来完成反转

代码如下：

#include <iostream>
    #include <algorithm>
    #include <vector>
    #include <stack>
    
    using namespace std;
    
    /*
    struct ListNode 
    {
         int val;
         ListNode *next;
         ListNode(int x) : val(x), next(NULL) {}
    };
    */
    
    class Solution 
    {
    public:
        ListNode* reverseBetween(ListNode* h, int m, int n)
        {
            if (h == NULL || h->next == NULL || m==n)
                return h;
            stack<ListNode*> sk;
            int count = 0;
            ListNode* head = new ListNode(-1);
            head->next = h;
            ListNode* pre = head;
            for (int i = 0; i < m - 1; i++)
                pre = pre->next;
    
            ListNode* start = pre->next;
            ListNode* then = start->next;
            for (int i = 0; i < n - m; i++)
            {
                start->next = then->next;
                then->next = pre->next;
                pre->next = then;
                then = start->next;
            }
            return head->next;
        }
    };