LeetCode 143. Reorder List

LeetCode 143. Reorder List 刷题笔记

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描述

Given a singly linked list L: L0?L1?…?Ln-1?Ln,
reorder it to: L0?Ln?L1?Ln-1?L2?Ln-2?…
You must do this in-place without altering the nodes’ values.

For example,Given {1,2,3,4}, reorder it to {1,4,2,3}.

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分析
题目要求我们把一个链表按照从头到尾,从尾到头分别每次取一个值交叉的排列成新的链表.
最直接的想法就是从中间分开链表,然后两头开始把后面的节点插入前半段链表.
但是由于是单链表,就算遍历得到最后的位置也不能向前迭代,所以要有一个把后半段链表翻转的过程,
于是得到如下思路:

1. 使用快慢指针法将链表分开为2部分
2. 后半段翻转
3. 2段合并

注:

快慢指针法:
使用两个指针slow和fast初始化都为链表头节点,
但是每次fast的步长为2即 fast = fast->next->next;
这样fast的速度为slow的2倍,当fast到了终点的时候slow在链表中部.

快慢指针可以解决多种链表问题,如判断链表是否有环路,有的话快慢指针一定会相遇,等等

AC代码

class Solution {public:    void reorderList(ListNode* head) {        if (head == NULL || !head->next || !head->next->next) return;        ListNode* slow = head;        ListNode* fast = head;        while(fast && fast->next) {           slow = slow->next;           fast = fast->next->next;        }         ListNode* subH = slow->next;        slow->next = NULL;        slow = NULL;        ListNode* pos = subH->next;        while(pos) {            subH->next = slow;            slow = subH;            subH = pos;            pos  = pos->next;        }        subH->next = slow;        slow = head;        ListNode * tem1, * tem2;        while(slow && subH) {            tem1 = slow->next;            tem2 = subH->next;            slow->next = subH;            subH->next = tem1;            subH = tem2;            slow = tem1;        }    }};