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≤1 \leq m \leq n \leq length of list.

分析

这题非常繁琐，有很多边界检查，15分钟内做到bug free很有难度！

代码

  1. // Reverse Linked List II
  2. // 迭代版，时间复杂度O(n)，空间复杂度O(1)
  3. class Solution {
  4. public:
  5.     ListNode *reverseBetween(ListNode *head, int m, int n) {
  6.         ListNode dummy(-1);
  7.         dummy.next = head;
  9.         ListNode *prev = &dummy;
  10.         for (int i = 0; i < m-1; ++i)
  11.             prev = prev->next;
  12.         ListNode* const head2 = prev;
  14.         prev = head2->next;
  15.         ListNode *cur = prev->next;
  16.         for (int i = m; i < n; ++i) {
  17.             prev->next = cur->next;
  18.             cur->next = head2->next;
  19.             head2->next = cur;  // 头插法
  20.             cur = prev->next;
  21.         }
  23.         return dummy.next;
  24.     }
  25. };

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/linear-list/linked-list/reverse-linked-list-ii.html