Reverse Linked List II LeetCode Solution

Problem – Reverse Linked List II LeetCode Solution

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 = [5], left = 1, right = 1
Output: [5]

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?

Reverse Linked List II LeetCode Solution in Java

public ListNode reverseBetween(ListNode head, int m, int n) {
    if(head == null) return null;
    ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list
    dummy.next = head;
    ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
    for(int i = 0; i<m-1; i++) pre = pre.next;
    
    ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
    ListNode then = start.next; // a pointer to a node that will be reversed
    
    // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
    // dummy-> 1 -> 2 -> 3 -> 4 -> 5
    
    for(int i=0; i<n-m; i++)
    {
        start.next = then.next;
        then.next = pre.next;
        pre.next = then;
        then = start.next;
    }
    
    // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
    // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)
    
    return dummy.next;
    
}

Reverse Linked List II LeetCode Solution in C++

class Solution {
public:
   ListNode* reverseBetween(ListNode* head, int m, int n) {
       ListNode *dummy = new ListNode(0), *pre = dummy, *cur;
       dummy -> next = head;
       for (int i = 0; i < m - 1; i++) {
           pre = pre -> next;
       }
       cur = pre -> next;
       for (int i = 0; i < n - m; i++) {
           ListNode* temp = pre -> next;
           pre -> next = cur -> next;
           cur -> next = cur -> next -> next;
           pre -> next -> next = temp;
       }
       return dummy -> next;
   }
};

Reverse Linked List II LeetCode Solution in Python

class Solution:
    # @param head, a ListNode
    # @param m, an integer
    # @param n, an integer
    # @return a ListNode
    def reverseBetween(self, head, m, n):
        if m == n:
            return head

        dummyNode = ListNode(0)
        dummyNode.next = head
        pre = dummyNode

        for i in range(m - 1):
            pre = pre.next
        
        # reverse the [m, n] nodes
        reverse = None
        cur = pre.next
        for i in range(n - m + 1):
            next = cur.next
            cur.next = reverse
            reverse = cur
            cur = next

        pre.next.next = cur
        pre.next = reverse

        return dummyNode.next
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