LeetCode(143) Reorder List

嘻唰唰对解答的分析很到位。

第一步：分析出链表分割点在第L/2 + 1个节点（第1，第2…)；

第二步：通过快慢指针找出链表的第L/2 + 1个节点，值得学习

[code]    二指针：前后指针、相向指针、背靠背指针、快慢指针
    三指针
    四指针

第三步：翻转链表

第四步：插入

选择将功能封装进子函数进行编程，降低编程的错误率以及降低程序出错后找错的难度。

[code]/**
 * 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) {

        if(head == NULL)

            return NULL;

        ListNode *fast = head;
        ListNode *slow = head;

        while(fast->next != NULL) {

            fast = fast->next;
            if(fast->next != NULL) fast = fast->next;
            slow = slow->next;

        }

        return slow;

    }

    ListNode *reverse(ListNode *head) {

        if(head == NULL || head->next == NULL)

            return head;

        ListNode *pre = head;    
        ListNode *cur = pre->next;

        while(cur) {

            ListNode *tmp = cur->next;
            cur->next = pre;

            pre = cur;
            cur = tmp;

        }

        head->next = NULL;

        return pre;

    }

    void reorderList(ListNode* head) {

        if(!head)

            return;

        ListNode *mid = findMid(head);
        ListNode *right = reverse(mid);

        ListNode *left = head;
        while(right->next != NULL) {

            ListNode *tmpRight = right->next;
            ListNode *tmpLeft = left->next;
            left->next = right;
            right->next = tmpLeft;

            left = left->next->next;
            right = tmpRight;

        }

    }

};