99. 重排链表

给定一个单链表L: L0→L1→…→Ln-1→Ln,

重新排列后为：L0→Ln→L1→Ln-1→L2→Ln-2→…

必须在不改变节点值的情况下进行原地操作。

您在真实的面试中是否遇到过这个题？

Yes

样例

给出链表 1->2->3->4->null，重新排列后为1->4->2->3->null

/**
 * Definition of ListNode
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *         this->val = val;
 *         this->next = NULL;
 *     }
 * }
 */


class Solution {
public:
    /*
     * @param head: The head of linked list.
     * @return: nothing
     */
    void reorderList(ListNode * head) {
        // write your code here
        	if (head == nullptr || head->next == nullptr)
        	{
        		return;
        	}
        
        	ListNode *slow = head, *fast = head->next;
        
        	while(fast&&fast->next)
        	{
        		slow = slow->next;
        		fast = fast->next->next;
        	}
        
        	fast = slow->next;
        	slow->next = nullptr;
        	ListNode *rHead = nullptr;
        	while (fast)
        	{
        		ListNode *r = fast->next;
        		fast->next = rHead;
        		rHead = fast;
        		fast = r;
        	}
        
        
        	fast = rHead;
        	slow = head;
        	while(slow&&fast)
        	{
        		ListNode *rr = fast->next;
        		ListNode *lr = slow->next;
        		fast->next = lr;
        		slow->next = fast;
        
        		fast = rr;
        		slow = lr;
        	}
    }
};