1.7交换链表中的相邻节点

交换链表中的相邻节点

题目描述：

把链表相邻元素翻转，例如给定链表为1——＞2一＞3一>4一＞5——>6一＞7，则翻转后的链表变为2一＞1一＞4一＞3一＞6一＞5一＞7

解题思路：

就地逆序法：

经过调整结点指针域的指向来直接调换相邻的两个结点。若是单链表刚好有偶数个结点，那么只须要将奇偶结点对调便可，若是链表有奇数个结点，那么只须要将除最后一个结点外的其它结点进行奇偶对调便可。

代码实现：

# -*-coding:utf-8-*- 
"""
@Author  : 图南
@Software: PyCharm
@Time    : 2019/9/6 18:34
"""
class Node:
    def __init__(self, data=None, next=None):
        self.data = data
        self.next = next


def print_link(head):
    if head is None or head.next is None:
        return None
    cur = head.next
    while cur.next != None:
        print(cur.data, end=' ')
        cur = cur.next
    print(cur.data)


def con_link(n):
    head = Node()
    cur = head
    for i in range(1, n+1):
        node = Node(i)
        cur.next = node
        cur = node
    return head


def reverseNode(head):
    pre = head
    cur = pre.next
    while cur is not None and cur.next is not None:
        next = cur.next
        pre.next = next
        cur.next = next.next
        next.next = cur
        pre = cur
        cur = pre.next
    return head


if __name__ == '__main__':
    head = con_link(6)
    print_link(head)
    head = reverseNode(head)
    print_link(head)

运行结果：