LeetCode 92. Reverse Linked List II

92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in one-pass.
将位置m的链接列表反转到n。一次通过。

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

代码

/*** 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) {int change_len=n-m+1;       //计算需要逆置的节点个数ListNode *pre_head=NULL;    //初始化开始逆置的节点的前驱ListNode *result=head;      //最终转换后的链表头节点，非特殊情况即为headwhile(head && --m){pre_head=head;      //记录head的前驱head=head->next;}ListNode *modify_list_tail = head;      //将modify_list_tail指向当前的head，即逆置后的链表尾ListNode *new_head = NULL;while(head&&change_len){//逆置change_len个节点ListNode *next=head->next;head->next=new_head;new_head=head;head=next;change_len--;       //每完成一个节点逆置，change_len--；}modify_list_tail->next=head;     //连接逆置后的链表尾与逆置段的后一个节点if(pre_head){//如果pre_head不空，说明不是从第一个节点开始逆置的，m>1pre_head->next=new_head;    //将逆置链表开始的节点前驱与逆置后的头节点连接}else{//如果pre_head为空，说明m==1，从第一个节点开始逆置，结果即为逆置后的头节点result=new_head;}return result;}
};

分析

首先找到四个关键的节点：逆置段头节点的前驱、逆置前头节点即逆置后尾节点、逆置前尾节点即逆置后头节点、逆置段尾节点的后继；

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LeetCode 92. Reverse Linked List II

92. Reverse Linked List II

代码

分析

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