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143. Reorder List

Description

You are given the head of a singly linked-list. The list can be represented as:

L0 → L1 → … → Ln - 1 → Ln

Reorder the list to be on the following form:

L0 → Ln → L1 → Ln - 1 → L2 → Ln - 2 → …

You may not modify the values in the list's nodes. Only nodes themselves may be changed.

 

Example 1:

Input: head = [1,2,3,4]
Output: [1,4,2,3]

Example 2:

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

 

Constraints:

  • The number of nodes in the list is in the range [1, 5 * 104].
  • 1 <= Node.val <= 1000

Solutions

Solution 1: Fast and Slow Pointers + Reverse List + Merge Lists

We first use fast and slow pointers to find the midpoint of the linked list, then reverse the second half of the list, and finally merge the two halves.

The time complexity is $O(n)$, where $n$ is the length of the linked list. The space complexity is $O(1)$.

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reorderList(self, head: Optional[ListNode]) -> None:
        fast = slow = head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next

        cur = slow.next
        slow.next = None

        pre = None
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        cur = head

        while pre:
            t = pre.next
            pre.next = cur.next
            cur.next = pre
            cur, pre = pre.next, t

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/**
 * 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 void reorderList(ListNode head) {
        ListNode fast = head, slow = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        ListNode cur = slow.next;
        slow.next = null;

        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        cur = head;

        while (pre != null) {
            ListNode t = pre.next;
            pre.next = cur.next;
            cur.next = pre;
            cur = pre.next;
            pre = t;
        }
    }
}

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode* head) {
        ListNode* fast = head;
        ListNode* slow = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }

        ListNode* cur = slow->next;
        slow->next = nullptr;

        ListNode* pre = nullptr;
        while (cur) {
            ListNode* t = cur->next;
            cur->next = pre;
            pre = cur;
            cur = t;
        }
        cur = head;

        while (pre) {
            ListNode* t = pre->next;
            pre->next = cur->next;
            cur->next = pre;
            cur = pre->next;
            pre = t;
        }
    }
};

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/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func reorderList(head *ListNode) {
    fast, slow := head, head
    for fast.Next != nil && fast.Next.Next != nil {
        slow, fast = slow.Next, fast.Next.Next
    }

    cur := slow.Next
    slow.Next = nil

    var pre *ListNode
    for cur != nil {
        t := cur.Next
        cur.Next = pre
        pre, cur = cur, t
    }
    cur = head

    for pre != nil {
        t := pre.Next
        pre.Next = cur.Next
        cur.Next = pre
        cur, pre = pre.Next, t
    }
}

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/**
 * 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)
 *     }
 * }
 */

/**
 Do not return anything, modify head in-place instead.
 */
function reorderList(head: ListNode | null): void {
    let slow = head;
    let fast = head;
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }

    let next = slow.next;
    slow.next = null;
    while (next) {
        [next.next, slow, next] = [slow, next, next.next];
    }

    let left = head;
    let right = slow;
    while (right.next) {
        const next = left.next;
        left.next = right;
        right = right.next;
        left.next.next = next;
        left = left.next.next;
    }
}

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// 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
//     }
//   }
// }
use std::collections::VecDeque;
impl Solution {
    pub fn reorder_list(head: &mut Option<Box<ListNode>>) {
        let mut tail = &mut head.as_mut().unwrap().next;
        let mut head = tail.take();
        let mut deque = VecDeque::new();
        while head.is_some() {
            let next = head.as_mut().unwrap().next.take();
            deque.push_back(head);
            head = next;
        }
        let mut flag = false;
        while !deque.is_empty() {
            *tail = if flag { deque.pop_front().unwrap() } else { deque.pop_back().unwrap() };
            tail = &mut tail.as_mut().unwrap().next;
            flag = !flag;
        }
    }
}

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/**
 * 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 {void} Do not return anything, modify head in-place instead.
 */
var reorderList = function (head) {
    let slow = head;
    let fast = head;
    while (fast.next && fast.next.next) {
        slow = slow.next;
        fast = fast.next.next;
    }

    let cur = slow.next;
    slow.next = null;

    let pre = null;
    while (cur) {
        const t = cur.next;
        cur.next = pre;
        pre = cur;
        cur = t;
    }
    cur = head;

    while (pre) {
        const t = pre.next;
        pre.next = cur.next;
        cur.next = pre;
        cur = pre.next;
        pre = t;
    }
};

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/**
 * 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 void ReorderList(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        ListNode cur = slow.next;
        slow.next = null;

        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        cur = head;

        while (pre != null) {
            ListNode t = pre.next;
            pre.next = cur.next;
            cur.next = pre;
            cur = pre.next;
            pre = t;
        }
    }
}

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