[Leetcode] 143. Reorder List 解题报告

题目：

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}.

思路：

思路很直观：1）将链表分成长度相等的两部分part1和part2；2）将part2翻转一下；3）将part1和翻转后的part2进行顺次合并。我第一次做这道题目的时候没有想到用two pointers，所以线性扫描了好几遍，虽然时间复杂度仍然符合题目要求，但是代码显得冗长。用了two pointers的思路之后，代码就简练多了。

代码：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    void reorderList(ListNode* head) {        if (!head || !head->next) {            return;        }        ListNode *slow = head, *fast = head;        ListNode *p = head, *q = head;        while (fast->next && fast->next->next) {            slow = slow->next;            fast = fast->next->next;        }        fast = slow->next;      // separate the list        slow->next = NULL;        p = fast, q = fast->next, fast->next = NULL;        while (q) {             // reverse the second list            ListNode *temp = q->next;            q->next = p;            p = q;            q = temp;        }        q = head;        while (p && q) {            ListNode *temp1 = q->next, *temp2 = p->next;            p->next = q->next;            q->next = p;            q = temp1, p = temp2;        }    }};