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143. 重排链表

题目描述

给定一个单链表 L 的头节点 head ，单链表 L 表示为：

L0 → L1 → … → Ln - 1 → Ln

请将其重新排列后变为：

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

不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

 

示例 1：

输入：head = [1,2,3,4]
输出：[1,4,2,3]

示例 2：

输入：head = [1,2,3,4,5]
输出：[1,5,2,4,3]

 

提示：

  • 链表的长度范围为 [1, 5 * 104]
  • 1 <= node.val <= 1000

解法

方法一：快慢指针 + 反转链表 + 合并链表

我们先用快慢指针找到链表的中点，然后将链表的后半部分反转，最后将左右两个链表合并。

时间复杂度 $O(n)$，其中 $n$ 是链表的长度。空间复杂度 $O(1)$。

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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:
        # 快慢指针找到链表中点
        fast = slow = head
        while fast.next and fast.next.next:
            slow = slow.next
            fast = fast.next.next

        # cur 指向右半部分链表
        cur = slow.next
        slow.next = None

        # 反转右半部分链表
        pre = None
        while cur:
            t = cur.next
            cur.next = pre
            pre, cur = cur, t
        cur = head

        # 此时 cur, pre 分别指向链表左右两半的第一个节点
        # 合并
        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) {
        // 快慢指针找到链表中点
        ListNode fast = head, slow = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        // cur 指向右半部分链表
        ListNode cur = slow.next;
        slow.next = null;

        // 反转右半部分链表
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        cur = head;

        // 此时 cur, pre 分别指向链表左右两半的第一个节点
        // 合并
        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) {
        // 快慢指针找到链表中点
        ListNode* fast = head;
        ListNode* slow = head;
        while (fast->next && fast->next->next) {
            slow = slow->next;
            fast = fast->next->next;
        }

        // cur 指向右半部分链表
        ListNode* cur = slow->next;
        slow->next = nullptr;

        // 反转右半部分链表
        ListNode* pre = nullptr;
        while (cur) {
            ListNode* t = cur->next;
            cur->next = pre;
            pre = cur;
            cur = t;
        }
        cur = head;

        // 此时 cur, pre 分别指向链表左右两半的第一个节点
        // 合并
        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) {
    // 快慢指针找到链表中点
    fast, slow := head, head
    for fast.Next != nil && fast.Next.Next != nil {
        slow, fast = slow.Next, fast.Next.Next
    }

    // cur 指向右半部分链表
    cur := slow.Next
    slow.Next = nil

    // 反转右半部分链表
    var pre *ListNode
    for cur != nil {
        t := cur.Next
        cur.Next = pre
        pre, cur = cur, t
    }
    cur = head

    // 此时 cur, pre 分别指向链表左右两半的第一个节点
    // 合并
    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;
    // 找到中心节点
    while (fast && fast.next) {
        slow = slow.next;
        fast = fast.next.next;
    }
    // 反转节点
    let next = slow.next;
    slow.next = null;
    while (next) {
        [next.next, slow, next] = [slow, next, next.next];
    }
    // 合并
    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) {
    // 快慢指针找到链表中点
    let slow = head;
    let fast = head;
    while (fast.next && fast.next.next) {
        slow = slow.next;
        fast = fast.next.next;
    }

    // cur 指向右半部分链表
    let cur = slow.next;
    slow.next = null;

    // 反转右半部分链表
    let pre = null;
    while (cur) {
        const t = cur.next;
        cur.next = pre;
        pre = cur;
        cur = t;
    }
    cur = head;

    // 此时 cur, pre 分别指向链表左右两半的第一个节点
    // 合并
    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) {
        // 快慢指针找到链表中点
        ListNode slow = head;
        ListNode fast = head;
        while (fast.next != null && fast.next.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        // cur 指向右半部分链表
        ListNode cur = slow.next;
        slow.next = null;

        // 反转右半部分链表
        ListNode pre = null;
        while (cur != null) {
            ListNode t = cur.next;
            cur.next = pre;
            pre = cur;
            cur = t;
        }
        cur = head;

        // 此时 cur, pre 分别指向链表左右两半的第一个节点
        // 合并
        while (pre != null) {
            ListNode t = pre.next;
            pre.next = cur.next;
            cur.next = pre;
            cur = pre.next;
            pre = t;
        }
    }
}

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