反转链表 II

• https://leetcode.cn/problems/reverse-linked-list-ii/description/

描述

• 给你单链表的头指针 head 和两个整数 left 和 right ，其中 left <= right
• 请你反转从位置 left 到位置 right 的链表节点，返回 反转后的链表

示例 1

输入：head = [1,2,3,4,5], left = 2, right = 4
输出：[1,4,3,2,5]

示例 2

输入：head = [5], left = 1, right = 1
输出：[5]

提示

• 链表中节点数目为 n
• 1 <= n <= 500
• -500 <= Node.val <= 500
• 1 <= left <= right <= n
• 进阶： 你可以使用一趟扫描完成反转吗？

Typescript 版算法实现

1 ) 方案1: 穿针引线

/**
 * 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)
 *     }
 * }
 */

const reverseLinkedList = (head: ListNode | null) => {
    let pre = null;
    let cur = head;

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

function reverseBetween(head: ListNode | null, left: number, right: number): ListNode | null {
    // 因为头节点有可能发生变化，使用虚拟头节点可以避免复杂的分类讨论
    const dummyNode = new ListNode(-1);
    dummyNode.next = head;

    let pre = dummyNode;
    // 第 1 步：从虚拟头节点走 left - 1 步，来到 left 节点的前一个节点
    // 建议写在 for 循环里，语义清晰
    for (let i = 0; i < left - 1; i++) {
        pre = pre.next;
    }

    // 第 2 步：从 pre 再走 right - left + 1 步，来到 right 节点
    let rightNode = pre;
    for (let i = 0; i < right - left + 1; i++) {
        rightNode = rightNode.next;
    }

    // 第 3 步：切断出一个子链表（截取链表）
    let leftNode = pre.next;
    let curr = rightNode.next;

    // 注意：切断链接
    pre.next = null;
    rightNode.next = null;

    // 第 4 步：同第 206 题，反转链表的子区间
    reverseLinkedList(leftNode);

    // 第 5 步：接回到原来的链表中
    pre.next = rightNode;
    leftNode.next = curr;
    return dummyNode.next;
};

2 ) 方案2: 一次遍历「穿针引线」反转链表（头插法）

/**
 * 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 reverseBetween(head: ListNode | null, left: number, right: number): ListNode | null {
    // 设置 dummyNode 是这一类问题的一般做法
    const dummy_node = new ListNode(-1);
    dummy_node.next = head;
    let pre = dummy_node;
    for (let i = 0; i < left - 1; ++i) {
        pre = pre.next;
    }

    let cur = pre.next;
    for (let i = 0; i < right - left; ++i) {
        const next = cur.next;
        cur.next = next.next;
        next.next = pre.next;
        pre.next = next;
    }
    return dummy_node.next;
};

3 ）方案3：局部反转法

/**
 * 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 reverseBetween(head: ListNode | null, left: number, right: number): ListNode | null {
    const dummy = {
        next: head
    }
    let tmp = dummy
    for (let i = 0; i < left - 1; i++) {
        tmp = tmp.next
    }

    let prev = tmp.next
    let cur = prev.next

    for (let j = 0; j < right - left; j++) {
        let next = cur.next
        cur.next = prev
        prev = cur
        cur = next // cur = cur.next
    }

    tmp.next.next = cur
    tmp.next = prev

    return dummy.next
};