【leetcode】链表类——翻转类题目

链表类——翻转类题目

序号	题目	难度	代码
25	Reverse Nodes in k-Group	Hard	python、java、c++
61	Rotate List	Medium	python、java、c++
92	Reverse Linked List II	Medium	python、java、c++
206	Reverse Linked List	Easy	python、java、c++

25. Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

Example:

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

Note:

• Only constant extra memory is allowed.
• You may not alter the values in the list's nodes, only nodes itself may be changed.

# 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:
        dummy = jump = ListNode(0)
        dummy.next = l = r = head
        while True:
            cnt = 0
            while r and cnt < k:
                r = r.next
                cnt += 1
            if cnt == k:
                pre, cur = r, l
                for _ in range(k):
                    cur.next, cur, pre = pre, cur.next,cur
                jump.next,jump,l = pre,l,r
            else:
                return dummy.next
            

61. Rotate List

Given a linked list, rotate the list to the right by k places, where k is non-negative.

Example 1:

Input: 1->2->3->4->5->NULL, k = 2

Output: 4->5->1->2->3->NULL

Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL

Example 2:

Input: 0->1->2->NULL, k = 4

Output: 2->0->1->NULL

Explanation:

rotate 1 steps to the right: 2->0->1->NULL

rotate 2 steps to the right: 1->2->0->NULL

rotate 3 steps to the right: 0->1->2->NULL

rotate 4 steps to the right: 2->0->1->NULL

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

class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        n, pre, current = 0 , None, head
        while current:
            pre,current = current,current.next
            n += 1
            
        if not n or not k % n:
            return head
        tail = head
        for _ in range(n - k % n - 1):
            tail = tail.next
        next, tail.next, pre.next = tail.next, None, head
        return next

92. Reverse Linked List II

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4

Output: 1->4->3->2->5->NULL

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

class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        dummy = pre = ListNode(0)
        dummy.next = head
        for _ in range(m-1):
            pre = pre.next
        cur= pre.next
        # reverse the defined part 
        node = None
        for _ in range(n-m+1):
            nxt = cur.next
            cur.next = node
            node = cur
            cur= nxt
        # connect three parts
        pre.next.next = cur
        pre.next = node
        return dummy.next       

206. Reverse Linked List

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL

Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

# 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 = None
        while head:
            pre, head.next,head = head,pre,head.next
        return pre