算法 中等 | 36. 翻转链表 II

• 题目描述
• 样例1
• 样例2
• 解题思路
• java题解
• C++题解
• python题解

题目描述

翻转链表中第m个节点到第n个节点的部分。

样例1

输入: 1->2->3->4->5->NULL, m = 2 and n = 4, 
输出: 1->4->3->2->5->NULL.

样例2

输入: 1->2->3->4->NULL, m = 2 and n = 3, 
输出: 1->3->2->4->NULL.

解题思路

1. 创建head.
2. 找到m的前一个节点-Pre
3. 记录Pre的下一个节点，它会是翻转链的尾部。
4. 翻转指定区间的链表，翻到最后一个节点时，把reverseTail.next指向它的next。这样就把翻转链表与之后的链表接起来了。
5. 返回dummynode的下一个节点。

java题解

public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (m >= n || head == null) {
            return head;
        }
        
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        head = dummy;
        
        for (int i = 1; i < m; i++) {
            if (head == null) {
                return null;
            }
            head = head.next;
        }
        
        ListNode premNode = head;
        ListNode mNode = head.next;
        ListNode nNode = mNode, postnNode = mNode.next;
        for (int i = m; i < n; i++) {
            if (postnNode == null) {
                return null;
            }
            ListNode temp = postnNode.next;
            postnNode.next = nNode;
            nNode = postnNode;
            postnNode = temp;
        }
        mNode.next = postnNode;
        premNode.next = nNode;
        
        return dummy.next;
    }
}

C++题解

using namespace std;

class Solution {
public:
    void reverse(ListNode *head) {
        ListNode *prev = NULL;
        while (head != NULL) {
            ListNode *temp = head->next;
            head->next = prev;
            prev = head;
            head = temp;
        }
    }

    ListNode *findkth(ListNode *head, int k) {
        for (int i = 0; i < k; i++) {
            if (head == NULL) {
                return NULL;
            }
            head = head->next;
        }
        return head;
    }

    ListNode *reverseBetween(ListNode *head, int m, int n) {
        ListNode *dummy = new ListNode(-1, head);
        ListNode *mth_prev = findkth(dummy, m - 1);
        ListNode *mth = mth_prev->next;
        ListNode *nth = findkth(dummy, n);
        ListNode *nth_next = nth->next;
        nth->next = NULL;

        reverse(mth);
        mth_prev->next = nth;
        mth->next = nth_next;
        return dummy->next;
    }
};

python题解

from lintcode import ListNode

class Solution:
    def reverse(self, head):
        prev = None
        while head != None:
            next = head.next
            head.next = prev
            prev = head
            head = next
        return prev

    def findkth(self, head, k):
        for i in xrange(k):
            if head is None:
                return None
            head = head.next
        return head

    def reverseBetween(self, head, m, n):
        dummy = ListNode(-1, head)
        mth_prev = self.findkth(dummy, m - 1)
        mth = mth_prev.next
        nth = self.findkth(dummy, n)
        nth_next = nth.next
        nth.next = None

        self.reverse(mth)
        mth_prev.next = nth
        mth.next = nth_next
        return dummy.next

• 点赞
• 收藏
• 分享
• • 文章举报

CN_wanku 发布了164 篇原创文章 · 获赞 23 · 访问量 1万+ 私信 关注

标签：II,head,ListNode,36,next,链表,nth,mth,prev
来源： https://blog.csdn.net/qq_43233085/article/details/104104978