[Leetcode] 143. Reorder List 解题报告_reorder list magicbean-优快云博客

本文介绍了一种链表重排序的方法，即将链表分为两部分，翻转第二部分后再与第一部分交替合并。文章通过示例展示了如何将链表{1,2,3,4}

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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;
        }
    }
};