剑指Offer 16. 合并两个排序的链表 （链表）

题目描述

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

题目地址

https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337?tpId=13&tqId=11169&rp=3&ru=%2Fta%2Fcoding-interviews&qru=%2Fta%2Fcoding-interviews%2Fquestion-ranking&tPage=1

思路

思路1：循环：如果两个链表不为空，进行比较，将小的赋给合并的指针头，小的链表走一步，合并链表走一步，如果有一个为空，跳出循环，并将另一不为空的链表后续部分赋给合并链表

思路2：递归：如果第一个链表为空，则返回第二个链表，如果第二个链表为空，则返回第一个链表，如果两个链表都为空，结果为空

两个链表都是排序好的，我们只需要从头遍历链表，判断当前指针，哪个链表的值小，即赋给合并链表指针。

Python

# -*- coding:utf-8 -*-
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

node1 = ListNode(1)
node2 = ListNode(3)
node3 = ListNode(5)
node1.next = node2
node2.next = node3
node4 = ListNode(2)
node5 = ListNode(4)
node6 = ListNode(6)
node4.next = node5
node5.next = node6

class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        # 循环
        # newHead = ListNode(-1)
        # pre = newHead
        # while pHead1 and pHead2:
        #     if pHead1.val < pHead2.val:
        #         pre.next = pHead1
        #         pHead1 = pHead1.next
        #     else:
        #         pre.next = pHead2
        #         pHead2 = pHead2.next
        #     pre = pre.next
        # pre.next = pHead1 if pHead1 else pHead2
        # return newHead.next

        # 递归
        if not pHead1:
            return pHead2
        if not pHead2:
            return pHead1
        newHead = None
        if pHead1.val < pHead2.val:
            newHead = pHead1
            newHead.next = self.Merge(pHead1.next, pHead2)
        else:
            newHead = pHead2
            newHead.next = self.Merge(pHead1, pHead2.next)
        return newHead


if __name__ == '__main__':
    result = Solution().Merge(node1,node4)
    print('合并链表：',end = ' ')
    while result:
        print(result.val,end = ' ')
        result = result.next