Suppose we have a linked list, we have to reverse it. So if the list is like 1 → 3 → 5 → 7, then the new reversed list will be 7 → 5 → 3 → 1
To solve this, we will follow this approach −
- Define one procedure to perform list reversal in a recursive way as to solve(head, back)
- if the head is not present, then return head
- temp := head.next
- head.next := back
- back = head
- if temp is empty, then return head
- head = temp
- return solve(head, back)
Example
Let us see the following implementation to get a better understanding −
class ListNode: def __init__(self, data, next = None): self.val = data self.next = next def make_list(elements): head = "ListNode(elements[0]) for element in elements[1:]: ptr = head while ptr.next: ptr = ptr.next ptr.next = ListNode(element) return head def print_list(head): ptr = head print('[', end = "") while ptr: print(ptr.val, end = ", ") ptr = ptr.next print(']') class Solution(object): def reverseList(self, head): """ :type head: ListNode :rtype: ListNode """ return self.solve(head,None) def solve(self, head, back): if not head: return head temp= head.next #print(head.val) head.next = back back = head if not temp: return head head = temp return self.solve(head,back) list1 = make_list([1,3,5,7]) ob1 = Solution() list2 = ob1.reverseList(list1) print_list(list2)
Input
list1 = [1,3,5,7]
Output
[7, 5, 3, 1, ]