LeetCode 143. Reorder List

最新推荐文章于 2022-03-07 13:49:13 发布

转载最新推荐文章于 2022-03-07 13:49:13 发布 · 44 阅读

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原文链接：http://www.cnblogs.com/yuanluo/p/7631988.html

本文介绍了一种链表重组算法，该算法将一个单向链表重新排列为特定形式，即首尾元素相邻，随后是剩余元素依次交错排列。文章详细展示了如何通过快慢指针找到中点，并反转后半部分链表，再将前后两部分链表交错连接起来的过程。

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Given a singly linked list L: L₀→L₁→…→L_n-1→L_n,
reorder it to: L₀→L_n→L₁→L_n-1→L₂→L_n-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}.

/**
 * 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 || !head->next->next) return;
        ListNode *fast = head;
        ListNode *slow = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;        
        }
        ListNode *mid = slow->next;
        slow->next = NULL;
        ListNode *curr = mid;
        ListNode *prev = NULL;
        
        while(curr) {
            ListNode *tmp = curr->next;
            curr->next = prev;
            prev = curr;
            curr = tmp;
        }
        
        while(head && prev) {
            ListNode *next = head->next;
            head->next = prev;
            prev = prev->next;
            head->next->next = next;
            head = next;
        }        
    }    
};