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Question

Formatted question description: https://leetcode.ca/all/206.html

206. Reverse Linked List

Level

Easy

Description

Reverse a singly linked list.

Example:

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

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

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

Solution

The recursive solution is to do the reverse operation for head.next, and set head.next.next = head and head.next = null.

The iterative solution uses two pointers, prev and curr. Initially, prev is null and curr points to head. Each step, let nextTemp be the next node of curr before reversing curr. Set curr.next = prev. After that, set both prev and curr to the next step (that is, prev = curr and curr = nextTemp). The loop ends when all nodes in the original list are reversed.

Code

  • 
public class Reverse_Linked_List {

    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) { val = x; }
     * }
     */

    /*
        1,2,3,4,5
        2,1 prev=2 - 3,4,5 current=3
        3,2,1 prev=3 - 4,5 current=4
        ...
     */
    // https://leetcode.com/problems/reverse-linked-list/solution/
    class Solution_oj_iterative {
        public ListNode reverseList(ListNode head) {
            ListNode prev = null;
            ListNode curr = head;
            while (curr != null) {
                ListNode nextTemp = curr.next;
                curr.next = prev;
                prev = curr;
                curr = nextTemp;
            }
            return prev;
        }
    }
    class Solution_oj_recursively {
        public ListNode reverseList(ListNode head) {
            if (head == null || head.next == null) return head;
            ListNode p = reverseList(head.next);
            head.next.next = head;
            head.next = null;
            return p;
        }
    }

    class Solution_recursively {
        public ListNode reverseList(ListNode head) {

            ListNode dummy = new ListNode(0);
            // dummy.next = head;

            ListNode current = head;

            reverse(dummy, current);

            return dummy.next;
        }

        private void reverse(ListNode dummy, ListNode current) {

            if (current == null) {
                return;
            }

            ListNode newHead = current.next;
            ListNode oldDummyNext = dummy.next;

            dummy.next = current;
            current.next = oldDummyNext;

            current = newHead;

            this.reverse(dummy, current);
        }
    }


    class Solution_iteratively {
        public ListNode reverseList(ListNode head) {

            ListNode dummy = new ListNode(0);
            // dummy.next = head;

            ListNode current = head;
            while (current != null) {

                ListNode newHead = current.next;
                ListNode oldDummyNext = dummy.next;

                dummy.next = current;
                current.next = oldDummyNext; // initial node, oldDummyNext is null, which is what we want, and which is why "// dummy.next = head;" is commented out above

                current = newHead;
            }

            return dummy.next;
        }
    }
}

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

/**
 * Definition for singly-linked list.
 * 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 reverseList(ListNode head) {
        ListNode dummy = new ListNode();
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = dummy.next;
            dummy.next = curr;
            curr = next;
        }
        return dummy.next;
    }
}

  • // OJ: https://leetcode.com/problems/reverse-linked-list/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode h;
        while (head) {
            auto p = head;
            head = head->next;
            p->next = h.next;
            h.next = p;
        }
        return h.next;
    }
};

  • # Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        pre, p = None, head
        while p:
            q = p.next
            p.next = pre
            pre = p
            p = q
        return pre


class Solution(object): # recursive
  def reverseList(self, root):
    if not root or not root.next:
      return root

    ret = self.reverseList(root.next)
    root.next.next = root # root.next is end of the newly reversed list ret
    root.next = None
    return ret


class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        dummy = ListNode()
        curr = head
        while curr:
            next = curr.next
            curr.next = dummy.next
            dummy.next = curr
            curr = next
        return dummy.next

  • /**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reverseList(head *ListNode) *ListNode {
	dummy := &ListNode{}
	curr := head
	for curr != nil {
		next := curr.Next
		curr.Next = dummy.Next
		dummy.Next = curr
		curr = next
	}
	return dummy.Next
}

  • /**
 * Definition for singly-linked list.
 * 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 reverseList(head: ListNode | null): ListNode | null {
    if (head == null) {
        return head;
    }
    let pre = null;
    let cur = head;
    while (cur != null) {
        const next = cur.next;
        cur.next = pre;
        [pre, cur] = [cur, next];
    }
    return pre;
}


  • /**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function (head) {
    let dummy = new ListNode();
    let curr = head;
    while (curr) {
        let next = curr.next;
        curr.next = dummy.next;
        dummy.next = curr;
        curr = next;
    }
    return dummy.next;
};


  • /**
 * Definition for singly-linked list.
 * 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 ReverseList(ListNode head) {
        ListNode pre = null;
        for (ListNode p = head; p != null;)
        {
            ListNode t = p.next;
            p.next = pre;
            pre = p;
            p = t;
        }
        return pre;
    }
}

  • // Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
impl Solution {
    pub fn reverse_list(head: Option<Box<ListNode>>) -> Option<Box<ListNode>> {
        match head {
            None => None,
            Some(mut head) => {
                let mut cur = head.next.take();
                let mut pre = Some(head);
                while let Some(mut node) = cur {
                    let next = node.next.take();
                    node.next = pre;
                    pre = Some(node);
                    cur = next;
                }
                pre
            }
        }
    }
}

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