# Reverse Linked List II

This is No.92 in leetcode.

The problem description:

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

I have a wrong idea on this problem, so I post the reference solution below:

``````class Solution {

// Object level variables since we need the changes
// to persist across recursive calls and Java is pass by value.
private boolean stop;
private ListNode left;

public void recurseAndReverse(ListNode right, int m, int n) {

// base case. Don't proceed any further
if (n == 1) {
return;
}

// Keep moving the right pointer one step forward until (n == 1)
right = right.next;

// Keep moving left pointer to the right until we reach the proper node
// from where the reversal is to start.
if (m > 1) {
this.left = this.left.next;
}

// Recurse with m and n reduced.
this.recurseAndReverse(right, m - 1, n - 1);

// In case both the pointers cross each other or become equal, we
// stop i.e. don't swap data any further. We are done reversing at this
// point.
if (this.left == right || right.next == this.left) {
this.stop = true;
}

// Until the boolean stop is false, swap data between the two pointers
if (!this.stop) {
int t = this.left.val;
this.left.val = right.val;
right.val = t;

// Move left one step to the right.
// The right pointer moves one step back via backtracking.
this.left = this.left.next;
}
}

public ListNode reverseBetween(ListNode head, int m, int n) {
this.left = head;
this.stop = false;
this.recurseAndReverse(head, m, n);
return head;
}
}``````

In this solution, it mainly use recursion to solve the problem. We will use two node, left and right to go over the linked list. At the beginning, we will check if we should move right further. And then we will check if we should move left. When the process of going over done, we could swap value of nodes.