Reverse Linked List II

描述

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->nullptr, m = 2 and n = 4,
return 1->4->3->2->5->nullptr.
Note: Given m, n satisfy the following condition: 1 <m< n < length of list.

分析

代码

// 迭代版，时间复杂度O(n)，空间复杂度O(1)

class Solution {public:    ListNode *reverseBetween(ListNode *head, int m, int n)     {        ListNode dummy(-1);        dummy.next = head;        ListNode* prev_m0 = &dummy;        for (int i = 1; i <= m-1; ++i)            prev_m0 = prev_m0 -> next; //指向第m-1个节点        listNode* prev_m = prev_m0->next; //指向第m个节点        for（int i = m+1; i <= n; ++i）        {            ListNode* psta=prev_m->next；//保存            prev_m->next = prev_m->next->next; //处理删除位置            psta->next = prev_m;//插入到第m-1个节点的后面            prev_m0->next = psta;                   }    }        return dummy.next;};