https://leetcode.com/problems/reverse-linked-list-ii/
92. Reverse Linked List II
- Total Accepted: 85657
- Total Submissions: 293208
- Difficulty: Medium
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,
return 1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
Code
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None
class Solution(object):
def reverseBetween(self, head, m, n):
"""
:type head: ListNode
:type m: int
:type n: int
:rtype: ListNode
"""
dummy = ListNode(3)
dummy.next = head
prev_t = dummy
for i in xrange(m-1):
prev_t = prev_t.next
rev_h = rev_t = prev_t.next
new_h = rev_h.next
for i in xrange(m, n):
new_h_next = new_h.next
new_h.next = rev_h
rev_h = new_h
new_h = new_h_next
prev_t.next = rev_h
rev_t.next = new_h
return dummy.next
Idea
I saw discussions on Leetcode that many people use several pointers as I do. So I believe my (iterative) solution is not too bad. Let me use a picture to illustrate prev_t, `rev_h`, `rev_t`, `new_h`:
