LenleDaytoy

剑指offer-25.合并两个排序列表

1.递归法

 1 class Solution:
 2     # 返回合并后列表
 3     def Merge(self, pHead1, pHead2):
 4         # write code here
 5         if pHead1 == None:
 6             return pHead2
 7         if pHead2 == None:
 8             return pHead1
 9         pMergedHead = None
10         if pHead1.val < pHead2.val:  #有递归，所以用if就可以比较依次所有元素，不用while
11             pMergedHead = pHead1
12             pMergedHead.next = self.Merge(pHead1.next, pHead2)  #pMergedHead可以将元素依次连接起来
13         else:
14             pMergedHead = pHead2
15             pMergedHead.next = self.Merge(pHead1, pHead2.next)
16         return pMergedHead

 

2.指针法（非递归）

class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        # write code here
        mergeHead = ListNode(90)  
        p = mergeHead
　　　　　while pHead1 and pHead2:
            if pHead1.val >= pHead2.val:
                mergeHead.next = pHead2
                pHead2 = pHead2.next
            else:
                mergeHead.next = pHead1
                pHead1 = pHead1.next
                  
            mergeHead = mergeHead.next   #写在if条件外，避免if和else里面重复写
        if pHead1:
            mergeHead.next = pHead1
        elif pHead2:
            mergeHead.next = pHead2
        return p.next

 

3.自己写的，代码不够简洁，

 1 class Solution:
 2     # 返回合并后列表
 3     def Merge(self, pHead1, pHead2):
 4         # write code here
 5         p1 = pHead1
 6         p2 = pHead2
 7         if not pHead1 and not pHead2:
 8             return None
 9         if not pHead1 and pHead2:   #不要忘记考虑一个链表为空的情况
10             return pHead2
11         if pHead1 and not pHead2:
12             return pHead1
13         if pHead1.val <= pHead2.val:
14             p = phead = pHead1
15             p1 = pHead1.next
16         else:
17             p = phead = pHead2
18             p2 = pHead2.next
19 
20         while  p1 and p2:
21             if p1.val <= p2.val:
22                 phead.next = p1
23                 p1 = p1.next
24             else:
25                 phead.next = p2
26                 p2 = p2.next
27             phead = phead.next
28         if p2:
29             phead.next = p2
30         if p1:
31             phead.next = p1
32         return p

 

posted on 2020-04-21 11:30  LenleDaytoy  阅读(120)  评论(0)  编辑  收藏  举报

导航