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剑指Offer之面试题25: 合并两个排序的链表
source link: https://blog.51cto.com/yuzhou1su/5141827
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合并两个有序链表
“Think ahead. Don't let day-to-day operations drive out planning.” — Donald Rumsfeld
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递增有序的规则。
示例1：
输入：1->2->4, 1->3->4
输出：1->1->2->3->4->4
输出：1->1->2->3->4->4
限制：
0 <= 链表长度 <= 1000
解题思路一：递归法
- 由于链表是升序排列，如果链表 1 的头结点小于链表 2 的头结点的值，那么链表 1 的头结点就是合并后链表的头结点。
- 并将下一层递归函数的返回值，链接到当前结点的尾部。
- 递归终止条件：至少一个为空，返回剩下的那个
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 == None:
return l2
if l2 == None:
return l1
if l1.val < l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 == None:
return l2
if l2 == None:
return l1
if l1.val < l2.val:
l1.next = self.mergeTwoLists(l1.next, l2)
return l1
else:
l2.next = self.mergeTwoLists(l1, l2.next)
return l2
解题思路二：双指针比较
分别用指针 l1, l2 来遍历两个链表，如果当前 l1 指向的数据小于 l2 指向的数据，则将 l1 指向的结点归入合并后的链表，否则将 l2 指向的结点归并到合并的链表中。
如果有一个链表遍历结束后，则把未结束的链表连接到合并链表后的链表尾部。
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 == None:
return l2
if l2 == None:
return l1
if l1.val >= l2.val:
head = l2
l2 = l2.next
else:
head = l1
l1 = l1.next
cur = head
while l1 and l2:
if l1.val <= l2.val:
cur.next = l1
cur = cur.next
l1 = l1.next
else:
cur.next = l2
cur = cur.next
l2 = l2.next
if l1 == None and l2:
cur.next = l2
elif l2 == None and l1:
cur.next = l1
return head
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
if l1 == None:
return l2
if l2 == None:
return l1
if l1.val >= l2.val:
head = l2
l2 = l2.next
else:
head = l1
l1 = l1.next
cur = head
while l1 and l2:
if l1.val <= l2.val:
cur.next = l1
cur = cur.next
l1 = l1.next
else:
cur.next = l2
cur = cur.next
l2 = l2.next
if l1 == None and l2:
cur.next = l2
elif l2 == None and l1:
cur.next = l1
return head
解题思路三：虚拟头结点
解题思想跟上述一样，但是为了减少对每一个节点的不同情况进行考虑，可以考虑建立一个虚拟头结点 dummy（这是一个很常用的链表题的技巧），然后用一个真正在走的结点 cur 指向这个 dummy ，每一个 cur 都会选取正确的之连接在以 dummy 为头结点的链表上。
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = cur = ListNode(0)
while l1 and l2:
if l1.val <= l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 if l1 else l2
return dummy.next
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = cur = ListNode(0)
while l1 and l2:
if l1.val <= l2.val:
cur.next = l1
l1 = l1.next
else:
cur.next = l2
l2 = l2.next
cur = cur.next
cur.next = l1 if l1 else l2
return dummy.next
时间复杂度: $$ O(m+n) $$，m,n 分别为链表 l1, l2 的长度
空间复杂度: $$ O(1) $$
感谢你的阅读，这里是不断学习中的yuzhou1su，keep coding, keep loving。
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