Leetcode-反转链表

206.逆转链表
Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL
方法一：迭代

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre, cur = None, head
        while cur:
                pre, cur, cur.next = cur, cur.next, pre
        return pre 

方法二：递归

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if not head or not head.next:
            return head
        p = self.reverseList(head.next)
        head.next.next = head
        head.next=None
        return p

24.两两交换链表
Given 1->2->3->4, you should return the list as 2->1->4->3.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def swapPairs(self, head: ListNode) -> ListNode:
        pre,pre.next = self,head
        while pre.next and pre.next.next:
            a = pre.next
            b = a.next
            a,b.next,b = b,a,b.next
            pre = a 
        return self.next

 

 

 

141.回环链表
Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where tail connects to the second node.

方法一：设置时间暴力循环
方法二：判断是否添加过该元素

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        l = []
        while head != None:
            if head.val in l:
                return True
            l.append(head)
            head=head.next
        return False       

方法三：双步运行

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        fir = sec =head
        while fir and sec and sec.next:
           fir = fir.next
           sec = sec.next.next
           if fir == sec:
                return True
        return False     

 

25. Reverse Nodes in k-Group

Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        if  head:
            node = head
            for _ in range(k-1):
                node = node.next
                if not node:
                    return head
            prev = self.reverseKGroup(node.next,k)
            
            while prev != node:
                prev, head.next, head = head, prev, head.next
            return prev