LEETCODE: Reorder List-优快云博客

本文详细阐述了如何重新排列单链表节点，使其形成交错排列的顺序，并提供了实现步骤和代码示例。

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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}.

写起来还是很麻烦的唉。。。。

/**
 * 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 == NULL || head->next == NULL) return;
        // Patition to two lists.
        ListNode *listone = head;
        ListNode *listtwo = head;
        ListNode *slow = head;
        ListNode *fast = head;
        int length = 0;
        while(slow != NULL) {
            slow = slow->next;
            length ++;
        }
        
        for(int ii = 0; ii < (length - 1) / 2; ii ++) {
            fast = fast->next;
        }
        listtwo = fast->next;
        fast->next = NULL;
        
        // Reverse listtwo.
        ListNode *pre = NULL;
        ListNode *current = listtwo;
        ListNode *after = listtwo->next;
        while(current != NULL && after != NULL) {
            current->next = pre;
            pre = current;
            current = after;
            after = after->next;
        }
        current->next = pre;
        listtwo = current;
        // Merge two lists.
        current = listone;
        while(listtwo != NULL) {
            // Insert listtwo after current.
            ListNode *temp = current->next;
            current->next = listtwo;
            current = temp;
            temp = listtwo->next;
            listtwo->next = current;
            listtwo = temp;
        }
    }
};