LeetCode题解92.reverse-linked-list-ii

题目地址

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

题目描述

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

思路

考虑取出需要反转的这一小段链表，反转完后再插入到原先的链表中。

以本题为例：

变换的是2,3,4这三个点，那么我们可以先取出2，用front指针指向2，然后当取出3的时候，我们把3加到2的前面，把front指针前移到3，依次类推，到4后停止，这样我们得到一个新链表4->3->2, front指针指向4。

对于原链表来说，有两个点的位置很重要，需要用指针记录下来，分别是1和5，把新链表插入的时候需要这两个点的位置。

用pre指针记录1的位置

当4结点被取走后，5的位置需要记下来

这样我们就可以把倒置后的那一小段链表加入到原链表中

LeetCode题解92.reverse-linked-list-ii

(图片来自： https://github.com/MisterBooo/LeetCodeAnimation)

关键点解析

  • 链表的基本操作（交换）
  • 虚拟节点dummy 简化操作
  • 考虑特殊情况 m 是 1 或者 n是链表长度的情况
  • 用四个变量记录特殊节点， 然后操作这四个节点使之按照一定方式连接即可。
    let midStartNode = null;
    let preMidStartNode = null;
    let midEndNode = null;
    let postMidEndNode = null;
  • 注意更新current和pre的位置， 否则有可能出现溢出

代码

语言支持：JS, C++, Python3

JavaScript Code:

/*
 * @lc app=leetcode id=92 lang=javascript
 *
 * [92] Reverse Linked List II
 *
 * https://leetcode.com/problems/reverse-linked-list-ii/description/
 *
 * algorithms
 * Medium (34.13%)
 * Total Accepted:    182.3K
 * Total Submissions: 532.8K
 * Testcase Example:  '[1,2,3,4,5]\n2\n4'
 *
 * Reverse a linked list from position m to n. Do it in one-pass.
 * 
 * Note: 1 ≤ m ≤ n ≤ length of list.
 * 
 * Example:
 * 
 * 
 * Input: 1->2->3->4->5->NULL, m = 2, n = 4
 * Output: 1->4->3->2->5->NULL
 * 
 * 
 */
/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} m
 * @param {number} n
 * @return {ListNode}
 */
var reverseBetween = function(head, m, n) {
    // 虚拟节点，简化操作
    const dummyHead = {
        next: head
    }

    let current = dummyHead.next; // 当前遍历的节点
    let pre = current; // 因为要反转，因此我们需要记住前一个节点
    let index = 0; // 链表索引，用来判断是否是特殊位置（头尾位置）

    // 上面提到的四个特殊节点
    let midStartNode = null;
    let preMidStartNode = null;
    let midEndNode = null;
    let postMidEndNode = null;

    while(current) {
        const next = current.next;
        index++;

        // 对 (m - n) 范围内的节点进行反转
        if (index > m && index <= n) {           
            current.next = pre;
        }

        // 下面四个if都是边界, 用于更新四个特殊节点的值
        if (index === m - 1) {
            preMidStartNode = current;
        }
        if (index === m) {
            midStartNode = current;
        }

        if (index === n + 1) {
            postMidEndNode = current;
        }

        if (index === n) {
            midEndNode = current;;
        }

        pre = current;

        current = next;
    }

    // 两个链表合并起来
    (preMidStartNode || dummyHead).next = midEndNode; // 特殊情况需要考虑
    midStartNode.next = postMidEndNode;

    return dummyHead.next;
};

C++ Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int s, int e) {
        if (s == e) return head;
        ListNode* prev = nullptr;
        auto cur = head;
        for (int i = 1; i < s; ++i) {
            prev = cur;
            cur = cur->next;
        }
        // 此时各指针指向：
        // x -> x -> x -> x  -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> x -> x -> x ->
        // ^head          ^prev ^cur
        ListNode* p = nullptr;
        auto c = cur;
        auto tail = c;
        ListNode* n = nullptr;
        for (int i = s; i <= e; ++i) {
            n = c->next;
            c->next = p;
            p = c;
            c = n;
        }
        // 此时各指针指向：
        // x -> x -> x -> x     8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1     x -> x -> x ->
        // ^head          ^prev ^p                                 ^cur  ^c
        //                                                         ^tail
        if (prev != nullptr) { // 若指向前一个节点的指针不为空，则说明s在链表中间（不是头节点）
            prev->next = p;
            cur->next = c;
            return head;
        } else {
            if (tail != nullptr) tail->next = c;
            return p;
        }
    }
};

Python Code:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
        """采用先翻转中间部分，之后与不变的前后部分拼接的思路"""
        # 处理特殊情况
        if m == n:
            return head
        
        # 例行性的先放一个开始节点，方便操作
        first = ListNode(0)
        first.next = head
        
        # 通过以下两个节点记录拼接点
        before_m = first  # 原链表m前的部分
        after_n = None  # 原链表n后的部分
        
        # 通过以下两个节点记录翻转后的链表
        between_mn_head = None
        between_mn_end = None
        
        index = 0
        cur_node = first
        while index < n:
            index += 1
            cur_node = cur_node.next
            if index == m - 1:
                before_m = cur_node
            elif index == m:
                between_mn_end = ListNode(cur_node.val)
                between_mn_head = between_mn_end
            elif index > m:
                temp = between_mn_head
                between_mn_head = ListNode(cur_node.val)
                between_mn_head.next = temp
                if index == n:
                    after_n = cur_node.next
        
        # 进行拼接
        between_mn_end.next = after_n
        before_m.next = between_mn_head
        
        return first.next
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