reverse_linked_list

View Source

 1from typing import Optional
 2from utils import ListNode
 3# @leet start
 4
 5
 6class Solution:
 7    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
 8        """
 9        To reverse a linked list, we want to keep two pointers, `prev` and `curr`
10        where `prev` refers to the previous pointer (or None) and `curr` refers to
11        the head of the linked list.
12        We then advance the linked list by taking the next and storing it in a temporary
13        variable, and then setting prev to curr and curr.next to prev (to reverse the list)
14        next, we set curr to the temporary variable and do that for every variable,
15        and at the end we return the prev pointer, which now holds the reversed list.
16        """
17        prev = None
18        curr = head
19        while curr:
20            next_temp = curr.next
21            prev = curr
22            curr.next = prev
23            curr = next_temp
24        return prev
25
26
27# @leet end
28
29
30def test():
31    assert 2 + 2 == 4

class Solution: View Source

 7class Solution:
 8    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
 9        """
10        To reverse a linked list, we want to keep two pointers, `prev` and `curr`
11        where `prev` refers to the previous pointer (or None) and `curr` refers to
12        the head of the linked list.
13        We then advance the linked list by taking the next and storing it in a temporary
14        variable, and then setting prev to curr and curr.next to prev (to reverse the list)
15        next, we set curr to the temporary variable and do that for every variable,
16        and at the end we return the prev pointer, which now holds the reversed list.
17        """
18        prev = None
19        curr = head
20        while curr:
21            next_temp = curr.next
22            prev = curr
23            curr.next = prev
24            curr = next_temp
25        return prev

def reverseList(self, head: Optional[utils.ListNode]) -> Optional[utils.ListNode]: View Source

 8    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
 9        """
10        To reverse a linked list, we want to keep two pointers, `prev` and `curr`
11        where `prev` refers to the previous pointer (or None) and `curr` refers to
12        the head of the linked list.
13        We then advance the linked list by taking the next and storing it in a temporary
14        variable, and then setting prev to curr and curr.next to prev (to reverse the list)
15        next, we set curr to the temporary variable and do that for every variable,
16        and at the end we return the prev pointer, which now holds the reversed list.
17        """
18        prev = None
19        curr = head
20        while curr:
21            next_temp = curr.next
22            prev = curr
23            curr.next = prev
24            curr = next_temp
25        return prev

To reverse a linked list, we want to keep two pointers, prev and curr where prev refers to the previous pointer (or None) and curr refers to the head of the linked list. We then advance the linked list by taking the next and storing it in a temporary variable, and then setting prev to curr and curr.next to prev (to reverse the list) next, we set curr to the temporary variable and do that for every variable, and at the end we return the prev pointer, which now holds the reversed list.

def test(): View Source

31def test():
32    assert 2 + 2 == 4