标签：merge ice list view style __name__ def 规则 ext
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
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
标签：merge ice list view style __name__ def 规则 ext
原文地址：https://www.cnblogs.com/huangqiancun/p/9784264.html
