Leetcode 143. Reorder List（Medium）

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}.

 搞了一天多，纠结到一个while 写成了if...........

思路很清晰，  找出链表的中点，翻转第二个链表，合并两个链表。

b 站视频讲解  

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* findMid(ListNode* head){   // 找出链表的中点，快慢指针解决，很经典。
        /*if (head == NULL || head -> next == NULL){
            return head;
        }*/
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast != NULL && fast -> next != NULL){    // 注意判断条件很容易出问题
            slow = slow -> next;
            fast = fast -> next -> next;
        }
        return slow;
    }
    ListNode* reverseList(ListNode* head){   // 翻转一个链表
        ListNode* newhead = NULL;
        while (head != NULL){                //  一开始写成了if 纠结了一天，长记性。。。
            ListNode* nextp = head -> next;
            head -> next = newhead;
            
            newhead =  head;   // 统一后移
            head = nextp;
        }
        return newhead;
    }
    ListNode* merge(ListNode* l1, ListNode* l2){   // 合并两个链表
        if (l2 == NULL){
            return l1;
        }
        ListNode* head = l1;
        
        while (l1 != NULL && l2 != NULL){
            ListNode* nextp = l1 -> next;
            l1 -> next = l2;
            l2 = l2 -> next;
            l1 -> next -> next = nextp;
            
            l1 =  nextp;     //  l1后移
        }
        return head;
    }
    void reorderList(ListNode* head) {
        if (head == NULL || head -> next == NULL){
            return;
        }
        ListNode* mid = findMid(head);   // 1.找到链表的中点 
        ListNode* l1 = head;
        ListNode* l2 = mid -> next;     
        mid -> next = NULL;              // 2. 断开链表
        l2 = reverseList(l2);            // 3. 翻转l2
        head = merge(l1, l2);            // 4. 合并两个链表
    }
};