# 143. Reorder List

Given a singly linked list L: L0→L1→…→Ln-1→Ln,

reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

# Solution

Approach 1: Iterate the list.

# Code (Python)

Approach 1:

# Code (C++)

Approach 1:

/**
 * 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) {
        // Reach the middle of the list.
        ListNode *slow = head;
        ListNode *fast = head;
        while (fast && fast->next)
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        // Reverse the last half of the list.
        ListNode *prev = slow;
        while (slow)
        {
            ListNode *next = slow->next;
            slow->next = prev;
            prev = slow;
            slow = next;
        }
        // Reorder the list.
        ListNode *tail = prev;
        ListNode dummyHead(0);
        dummyHead.next = head;
        ListNode *curr = &dummyHead;
        while (head != tail)
        {
            curr->next = head;
            head = head->next;
            curr = curr->next;
            curr->next = tail;
            tail = tail->next;
            curr = curr->next;
        }
        curr->next = head;
        curr = curr->next;
        if (curr)
            curr->next = NULL;
    }
};