Leetcode解题-链表(2.2.2)ReverseLinkedList

题目：2.2.2 Reverse Linked List II

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.

链表旋转是个经典问题，可以锻炼我们操作链表的能力，值得好好学习一下。

我的方法：对head和tail特殊化处理

当m<=i<=n时，通过记录前继结点prev，简单的调转next指针方向，并记录m结点pm。最后迭代到n+1时，再处理first（pm->next）和last（p）结点的指向问题。

更好的方法：一般化处理

将整个过程一般化处理的方法为：不断地将当前结点cur插入到head的后面。（见下面图示）第一次将3插入到1后面，第二次将4插入到1后面，即1=>2=>3=>4旋转为1=>3=>2=>4。如果能解决的话，那么继续1=>3=>2=>4=>5就能旋转为1=>4=>3=>2=>5，从而解决问题。总结：首先举最简单的例子解决，然后看解决办法能否继续推广到更复杂一些的例子，避免复杂的特殊化处理。

代码实现

理解了整体设计，自己实现或读代码就很简单了。开始真正处理前要记住head（m的前一个结点）和pm（索引为m的结点），就可以开始旋转了。

时间： 2024-10-19 07:04:09

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