Reorder List

题目143：Reorder List

Reorder List Total Accepted: 46551 Total Submissions: 221280 My Submissions Question Solution
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}.
思路分析：
先用快慢指针找到链表的中点，将链表的后半部分转置，转置后合并这两个半个链表。
常用到的操作：
1. 快慢指针找链表中点；
2. 转置一个链表(递归和非递归实现)；
3. 链表的插入、删除

class Solution {
public:
    ListNode *reverse(ListNode* head) {
        if (0 == head || 0 == head->next)
            return head;
        ListNode *pre, *cur;
        pre = 0;
        cur = head;
        while (cur) {
            ListNode *pnext = cur->next;
            cur->next = pre;
            pre = cur;
            cur = pnext;
        }
        return pre;
    }
    void reorderList(ListNode* head) {
        if (head == 0 || head->next == 0)
            return;
        ListNode *fast, *slow, *mid;
        fast = head;
        slow = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        /* 链表的一半是slow的next */
        mid = slow->next;
        slow->next = 0;
        ListNode *reverse_head, *merge_head;
        reverse_head = reverse(mid);
        fast = head;
        /* 可以这么合并oo */
        while (fast && reverse_head) {
            ListNode *temp = reverse_head->next;
            reverse_head->next = fast->next;
            fast->next = reverse_head;
            fast = reverse_head->next;
            reverse_head = temp;
        }
        /* 这里两个子链表是转置的子链表空，则合并就结束了。 */
    }
};