# Question

Formatted question description: https://leetcode.ca/all/92.html

Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list.

Example 1: Input: head = [1,2,3,4,5], left = 2, right = 4
Output: [1,4,3,2,5]


Example 2:

Input: head = , left = 1, right = 1
Output: 


Constraints:

• The number of nodes in the list is n.
• 1 <= n <= 500
• -500 <= Node.val <= 500
• 1 <= left <= right <= n

Follow up: Could you do it in one pass?

# Algorithm

Take the example in the title, the three points that are transformed are 2, 3, and 4. We need to find the first node before the transformation node, as long as the pre is moved backward by m-1 steps.

Why do you want to subtract 1, because the problem is counting from 1, here only one step is taken, which is node 1, and use pre to point to it. What if the node 1 starts to change? This is why we use the dummy node. Pre can also point to the dummy node. Then the exchange is about to begin. Since only two nodes can be exchanged at a time, we follow the exchange sequence as follows:

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

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

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


This problem is a follow-up problem for problem 206. In this problem, only the nodes from position m to n should be reversed, and the reverse operation should be done in one-pass. The iterative solution for problem 206 can be adapted here, with some modification.

Obviously, if the list is null, or there is only one element in the list, or position m is the same as position n, then no reverse needs to be done and just return the original list.

Since the head may also be reversed, create a dummyHead whose next pointer points to head. Find the reverse start node, the reverse end node, the last node before reverse and the first node after reverse, and then do the reverse operation.

For example, for linked list 1->2->3->4->5->NULL, m = 2, n = 4, the reverse start node is 2, the reverse end node is 4, the last node before reverse is 1, and the first node after reverse is 5.

Like the iterative solution for problem 206, use two pointers prev and curr to do the iteration. After the loop, set the last node before reverse to point to the reverse end node (which is the start of the reverse part after reverse), and set the reverse start node (which is the end of the reverse part after reverse) to point to the first node after reverse.

Finally, return dummyHead.next.

# Code

• import java.util.Stack;

/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/

public class Solution_optimize {
public ListNode reverseBetween(ListNode head, int m, int n) {

ListNode dummy = new ListNode(0);
ListNode prev = dummy;

// I missed this one

for (int i = 1; i < m; i++) {
prev = p;
p = p.next;
}

// @note:@memorize: now p is pointing to m-th node
ListNode originalFirst = p;
for (int i = m; i < n; i++) {

// swap
}

return dummy.next;
}
}

public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {

if (m > n) {
}

int diff = n - m;
if (diff == 0) {
}

// set diff for both pointers
// @note: corner case: n-m > list-length
while (diff > 0 && p1 != null) {
p1 = p1.next;
diff--;
}

ListNode dummy = new ListNode(0);

ListNode prev = dummy;

int mm = m;
while (mm - 1> 0) {
prev = prev.next;
p1 = p1.next;
p2 = p2.next;

mm--;
}

ListNode nextRecord = p1.next;
diff = n - m;

// start reverse from p1 to p2
// 1->2->3->4->5
Stack<ListNode> sk = new Stack<>();
while (p2 != p1.next) { // @note: here, when p1==p2 should enter loop
sk.push(p2);
p2 = p2.next;
}

while (!sk.isEmpty()) {
prev.next = sk.pop();
prev = prev.next;
}

prev.next = nextRecord;

return dummy.next;
}
}
}

############

/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode() {}
*     ListNode(int val) { this.val = val; }
*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode reverseBetween(ListNode head, int left, int right) {
if (head.next == null || left == right) {
}
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
for (int i = 0; i < left - 1; ++i) {
pre = pre.next;
}
ListNode p = pre;
ListNode q = pre.next;
ListNode cur = q;
for (int i = 0; i < right - left + 1; ++i) {
ListNode t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
p.next = pre;
q.next = cur;
return dummy.next;
}
}

• // OJ: https://leetcode.com/problems/reverse-linked-list-ii/
// Time: O(N)
// Space: O(1)
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode dummy, *p = &dummy;
for (int i = 1; i < m; ++i) p = p->next;
auto q = p->next, tail = q;
for (int i = m; i <= n; ++i) {
auto node = q;
q = q->next;
node->next = p->next;
p->next = node;
}
tail->next = q;
return dummy.next;
}
};

• # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
def reverseBetween(
self, head: Optional[ListNode], left: int, right: int
) -> Optional[ListNode]:
if head.next is None or left == right:
pre = dummy
for _ in range(left - 1):
pre = pre.next
p, q = pre, pre.next
cur = q
# +1, so when for loop done, 'cur' is at right's next node
for _ in range(right - left + 1):
t = cur.next
cur.next = pre
pre, cur = cur, t
p.next = pre
# p,q did not change by the for loop, but now pre at right(or called m) node, cur at right+1 node
q.next = cur
return dummy.next

############

class Solution_optimize:
def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
dummy = ListNode(0)
prev = dummy

for i in range(1, m):
prev = p
p = p.next

original_first = p # this is anchor node always pointing to the next-to-be-swapped
for i in range(m, n):

return dummy.next

############

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
"""
:type m: int
:type n: int
:rtype: ListNode
"""

def reverse(root, prep, k):
cur = root
pre = None
next = None
while cur and k > 0:
next = cur.next
cur.next = pre
pre = cur
cur = next
k -= 1
root.next = next
prep.next = pre
return pre

dummy = ListNode(-1)
k = 1
p = dummy
start = None
while p:
if k == m:
start = p
if k == n + 1:
reverse(start.next, start, n - m + 1)
return dummy.next
k += 1
p = p.next


• /**
* type ListNode struct {
*     Val int
*     Next *ListNode
* }
*/
func reverseBetween(head *ListNode, left int, right int) *ListNode {
if head.Next == nil || left == right {
}
pre := dummy
for i := 0; i < left-1; i++ {
pre = pre.Next
}
p, q := pre, pre.Next
cur := q
for i := 0; i < right-left+1; i++ {
t := cur.Next
cur.Next = pre
pre = cur
cur = t
}
p.Next = pre
q.Next = cur
return dummy.Next
}

• /**
* class ListNode {
*     val: number
*     next: ListNode | null
*     constructor(val?: number, next?: ListNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.next = (next===undefined ? null : next)
*     }
* }
*/

function reverseBetween(
left: number,
right: number,
): ListNode | null {
const n = right - left;
if (n === 0) {
}

const dummy = new ListNode(0, head);
let pre = null;
let cur = dummy;
for (let i = 0; i < left; i++) {
pre = cur;
cur = cur.next;
}
const h = pre;
pre = null;
for (let i = 0; i <= n; i++) {
const next = cur.next;
cur.next = pre;
pre = cur;
cur = next;
}
h.next.next = cur;
h.next = pre;
return dummy.next;
}


• /**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {number} left
* @param {number} right
* @return {ListNode}
*/
var reverseBetween = function (head, left, right) {
if (!head.next || left == right) {
}
const dummy = new ListNode(0, head);
let pre = dummy;
for (let i = 0; i < left - 1; ++i) {
pre = pre.next;
}
const p = pre;
const q = pre.next;
let cur = q;
for (let i = 0; i < right - left + 1; ++i) {
const t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
p.next = pre;
q.next = cur;
return dummy.next;
};


• /**
* public class ListNode {
*     public int val;
*     public ListNode next;
*     public ListNode(int val=0, ListNode next=null) {
*         this.val = val;
*         this.next = next;
*     }
* }
*/
public class Solution {
public ListNode ReverseBetween(ListNode head, int left, int right) {
if (head.next == null || left == right) {
}
ListNode dummy = new ListNode(0, head);
ListNode pre = dummy;
for (int i = 0; i < left - 1; ++i) {
pre = pre.next;
}
ListNode p = pre;
ListNode q = pre.next;
ListNode cur = q;
for (int i = 0; i < right - left + 1; ++i) {
ListNode t = cur.next;
cur.next = pre;
pre = cur;
cur = t;
}
p.next = pre;
q.next = cur;
return dummy.next;
}
}