#Leetcode# 143. Reorder List

https://leetcode.com/problems/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.

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 || !head -> next -> next) return ;
        ListNode *slow = head; ListNode *fast = head;
        while(fast -> next && fast -> next -> next) {
            slow = slow -> next;
            fast = fast -> next -> next;
        }
        
        ListNode *dummy = slow -> next;
        slow -> next = NULL;

        dummy = reverseList(dummy);
        
        while(head && dummy) {
            ListNode *New = head -> next;
            head -> next = dummy;
            dummy = dummy -> next;
            head -> next -> next = New;
            head = New;
        }
    }
    
    ListNode *reverseList(ListNode *head) {
        if(!head || !head -> next) return head;
        ListNode *New = reverseList(head -> next);
        head -> next -> next = head;
        head -> next = NULL;
        return New;
    }
};

　　快慢指针把链表从中间分开然后反转链表 之后把两个链表合并 希望今天出门剪头发之前 Leetcode 刷到 200