🌵92. Reverse Linked List II
https://leetcode.com/problems/reverse-linked-list-ii/
js
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @param {number} m
* @param {number} n
* @return {ListNode}
*/
var reverseBetween = function(head, m, n) {
var dummy = new ListNode()
dummy.next = head
var tail = dummy
for (let i = 1; i < m; i++) {
tail = tail.next
}
var revHead = tail.next
for (let i = m; i < n; i++) {
let temp = revHead.next
revHead.next = temp.next
temp.next = tail.next
tail.next = temp
}
return dummy.next
}py
# 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
"""
tail = dummy = ListNode(None)
dummy.next = head
for i in range(1, m):
tail = tail.next
revHead = tail.next
for i in range(m, n):
temp = revHead.next
revHead.next = temp.next
temp.next = tail.next
tail.next = temp
return dummy.next