143. 重排链表

给定一个单链表 LL0→L1→…→Ln-1→Ln ,

将其重新排列后变为: L0→LnL1→Ln-1→L2→Ln-2→…

不能只是单纯的改变节点内部的值,而是需要实际的进行节点交换。

示例1:给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例2:给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

class Solution(object):
    def reorderList(self, head):
        if not head:
            return head
        slow = head
        fast = head       
        #利用快慢指针找到中点(链表中经常用到"快慢指针"种方法,应掌握!!!)
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        #slow为找到的中点,将slow后的链表进行反转为rev链表
        rev = self.reverse(slow.next)
        slow.next = None #切断链表,原链表到slow就结束了
        #将反转的链表rev中的每一项项插入原链表
        cur = head
        while cur and rev:
            temp1 = cur.next
            temp2 = rev.next
            cur.next = rev
            rev.next = temp1
            rev = temp2
            cur = temp1    
        def reverse(self, head):  #反转链表(基本操作,要熟练写出!!!)
           pre = None
           cur = head
           while cur:
               lat = cur.next
               cur.next = pre
               pre = cur
               cur = lat
           return pre
#利用栈来实现,代码量较少,但是leetcode上用时大
if not head or not head.next:
            return head
        cur = head
        stack = []
        while cur:
            stack.append(cur)
            cur = cur.next
        cur = stack.pop(0)  #栈底
        while stack:
            cur.next = stack.pop()  #栈顶
            cur = cur.next
            if stack:
                cur.next = stack.pop(0)
                cur=cur.next
        cur.next = None
posted @ 2019-04-25 11:11  USTC丶ZCC  阅读(246)  评论(0编辑  收藏  举报