Reverse Linked List II(旋转链表的指定部分)

Leetcode

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.

旋转链表的有序部分,思路：

1. 申请一个辅助链表，将第m到第n个节点依次插入到辅助链表头结点后，完成异地置逆。
2. 将辅助链表插回到原链表的 m 到 n 之间。

/**
 * 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) {
       if(head == nullptr || m >= n)
            return head;
       ListNode * newHead = new ListNode(0); //原链表前增加头结点
       newHead->next = head;
       ListNode * tempHead = new ListNode(0); //辅助链表的头结点
       ListNode * pre = newHead; //需要置逆的前一个节点
       int i = 1; //第i个节点，永远指向需要被删除的那么节点
       while(pre->next != nullptr && i < m) //将i指向第m个节点，pre指向第m-1个节点
       {
            pre = pre->next;
            i++;
       }
       //这里是否需要判断没有到第m个节点?
       if(i < m) return head;
       ListNode * p = nullptr;
       while(i <= n && pre->next != nullptr)
       {
            p = pre->next;
            pre->next = p->next; //在原链表上去掉第i个节点
            p->next = tempHead->next;
            tempHead->next = p; //将第i个节点添加到辅助链表上
            i++;
       }
       p = tempHead;
       while(p->next != nullptr) //p指向辅助链表的最后一个节点
            p = p->next;
        p->next = pre->next;
        pre->next = tempHead->next; //将辅助链表除头结点以外部分插入到原链表pre后面
        return newHead->next;
    }
};

(每日算法)LeetCode --- Reverse Linked List II(旋转链表的指定部分)