代码随想录算法训练营第四天|24. 两两交换链表中的节点、19. 删除链表的倒数第 N 个结点、面试题 02.07. 链表相交、142. 环形链表 II

wuyijia / 2023-05-14 / 原文

24. 两两交换链表中的节点

【注意】

1.操作指针一定要指向要反转两个结点的前一个结点

2.遍历链表的时候什么时候终止，cur.next.next == None ,则遍历结束（n为奇数），cur.next == None(n为偶数)。

3.时间复杂度O(n),空间复杂度O(1)

【代码】

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 #     def __init__(self, val=0, next=None):
 4 #         self.val = val
 5 #         self.next = next
 6 class Solution(object):
 7     def swapPairs(self, head):
 8         """
 9         :type head: ListNode
10         :rtype: ListNode
11         """
12         dumyhead = ListNode(next = head)
13         cur = dumyhead
14 
15         while cur.next and cur.next.next: #顺序不能变，否则会发生空指针异常
16             firstnode = cur.next #临时保存第一个结点
17             thirdnode = cur.next.next.next #临时保存第三个结点
18             cur.next = cur.next.next #步骤1
19             cur.next.next = firstnode #步骤2
20             firstnode.next = thirdnode #步骤3
21             cur = cur.next.next
22 
23         return dumyhead.next

19. 删除链表的倒数第 N 个结点

【代码】

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 #     def __init__(self, val=0, next=None):
 4 #         self.val = val
 5 #         self.next = next
 6 class Solution(object):
 7     def removeNthFromEnd(self, head, n):
 8         """
 9         :type head: ListNode
10         :type n: int
11         :rtype: ListNode
12         """
13         dumyhead = ListNode(next = head)
14         fast = dumyhead
15         slow = dumyhead
16         #n += 1 #防止fast为空，发生空指针异常  #快指针走n+1步
17         while n >= 0 and fast:
18             fast = fast.next
19             n -= 1
20         while fast:
21             slow = slow.next
22             fast = fast.next #慢指针才能到要操作结点的前一个结点
23         slow.next = slow.next.next #遍历结束之后删除
24         
25         return dumyhead.next

面试题 02.07. 链表相交

【注意】

1.交点不是数值相等，而是指针相等。

【代码】

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 #     def __init__(self, x):
 4 #         self.val = x
 5 #         self.next = None
 6 
 7 class Solution(object):
 8     def getIntersectionNode(self, headA, headB):
 9         """
10         :type head1, head1: ListNode
11         :rtype: ListNode
12         """
13         lenA, lenB = 0, 0
14         cur = headA
15 
16         while cur: #求链表A长度
17             cur = cur.next
18             lenA += 1
19         
20         cur = headB
21         while cur: #求链表B长度
22             cur = cur.next
23             lenB += 1
24 
25         curA, curB = headA, headB
26         if lenA > lenB:  # 让curB为最长链表的头，lenB为其长度
27             curA, curB = curB, curA
28             lenA, lenB = lenB, lenA
29 
30         for _ in range(lenB - lenA): # 让curA和curB在同一起点上（末尾位置对齐）
31              curB = curB.next 
32 
33         while curA:
34             if curA == curB:
35                 return curA
36             else:
37                 curA = curA.next
38                 curB = curB.next
39         return None