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feat: add python and java solutions to leetcode problem: No.0092
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‎README.md

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@@ -62,6 +62,7 @@
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1. [合并 K 个排序链表](/solution/0000-0099/0023.Merge%20k%20Sorted%20Lists/README.md)
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1. [排序链表](/solution/0100-0199/0148.Sort%20List/README.md)
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1. [反转链表](/solution/0200-0299/0206.Reverse%20Linked%20List/README.md)
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1. [反转链表 II](/solution/0000-0099/0092.Reverse%20Linked%20List%20II/README.md)
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1. [重排链表](/solution/0100-0199/0143.Reorder%20List/README.md)
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1. [旋转链表](/solution/0000-0099/0061.Rotate%20List/README.md)
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1. [回文链表](/solution/0200-0299/0234.Palindrome%20Linked%20List/README.md)

‎README_EN.md

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@@ -60,6 +60,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
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1. [Merge k Sorted Lists](/solution/0000-0099/0023.Merge%20k%20Sorted%20Lists/README_EN.md)
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1. [Sort List](/solution/0100-0199/0148.Sort%20List/README_EN.md)
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1. [Reverse Linked List](/solution/0200-0299/0206.Reverse%20Linked%20List/README_EN.md)
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1. [Reverse Linked List II](/solution/0000-0099/0092.Reverse%20Linked%20List%20II/README_EN.md)
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1. [Reorder List](/solution/0100-0199/0143.Reorder%20List/README_EN.md)
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1. [Rotate List](/solution/0000-0099/0061.Rotate%20List/README_EN.md)
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1. [Palindrome Linked List](/solution/0200-0299/0234.Palindrome%20Linked%20List/README_EN.md)

‎solution/0000-0099/0092.Reverse Linked List II/README.md

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@@ -26,15 +26,70 @@
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
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if head is None or head.next is None or m == n:
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return head
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dummy = ListNode(0)
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dummy.next = head
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pre, cur = dummy, head
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for _ in range(m - 1):
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pre = cur
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cur = cur.next
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p, q = pre, cur
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for _ in range(n - m + 1):
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t = cur.next
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cur.next = pre
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pre = cur
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cur = t
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p.next = pre
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q.next = cur
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return dummy.next
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode reverseBetween(ListNode head, int m, int n) {
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if (head == null || head.next == null || m == n) {
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return head;
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}
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ListNode dummy = new ListNode(0);
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dummy.next = head;
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ListNode pre = dummy, cur = head;
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for (int i = 0; i < m - 1; ++i) {
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pre = cur;
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cur = cur.next;
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}
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ListNode p = pre, q = cur;
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for (int i = 0; i < n - m + 1; ++i) {
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ListNode t = cur.next;
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cur.next = pre;
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pre = cur;
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cur = t;
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}
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p.next = pre;
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q.next = cur;
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return dummy.next;
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}
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}
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```
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### **...**

‎solution/0000-0099/0092.Reverse Linked List II/README_EN.md

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@@ -25,13 +25,68 @@
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### **Python3**
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
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if head is None or head.next is None or m == n:
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return head
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dummy = ListNode(0)
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dummy.next = head
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pre, cur = dummy, head
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for _ in range(m - 1):
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pre = cur
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cur = cur.next
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p, q = pre, cur
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for _ in range(n - m + 1):
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t = cur.next
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cur.next = pre
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pre = cur
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cur = t
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p.next = pre
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q.next = cur
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return dummy.next
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```
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### **Java**
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```java
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode reverseBetween(ListNode head, int m, int n) {
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if (head == null || head.next == null || m == n) {
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return head;
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}
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ListNode dummy = new ListNode(0);
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dummy.next = head;
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ListNode pre = dummy, cur = head;
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for (int i = 0; i < m - 1; ++i) {
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pre = cur;
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cur = cur.next;
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}
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ListNode p = pre, q = cur;
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for (int i = 0; i < n - m + 1; ++i) {
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ListNode t = cur.next;
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cur.next = pre;
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pre = cur;
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cur = t;
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}
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p.next = pre;
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q.next = cur;
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return dummy.next;
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}
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}
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```
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### **...**

‎solution/0000-0099/0092.Reverse Linked List II/Solution.java

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@@ -1,25 +1,32 @@
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode reverseBetween(ListNode head, int m, int n) {
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ListNode dummy = new ListNode(-1);
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if (head == null || head.next == null || m == n) {
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return head;
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}
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ListNode dummy = new ListNode(0);
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dummy.next = head;
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ListNode pre = dummy;
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int i = 0;
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for (; i < m - 1; ++i) {
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pre = pre.next;
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ListNode pre = dummy, cur = head;
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for (int i = 0; i < m - 1; ++i) {
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pre = cur;
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cur = cur.next;
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}
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11-
ListNode head2 = pre;
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pre = pre.next;
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ListNode cur = pre.next;
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for (; i < n - 1; ++i) {
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pre.next = cur.next;
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cur.next = head2.next;
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head2.next = cur;
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cur = pre.next;
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ListNode p = pre, q = cur;
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for (int i = 0; i < n - m + 1; ++i) {
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ListNode t = cur.next;
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cur.next = pre;
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pre = cur;
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cur = t;
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}
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p.next = pre;
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q.next = cur;
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return dummy.next;
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}
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}

‎solution/0000-0099/0092.Reverse Linked List II/Solution.py

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@@ -0,0 +1,25 @@
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
6+
7+
class Solution:
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def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode:
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if head is None or head.next is None or m == n:
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return head
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dummy = ListNode(0)
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dummy.next = head
13+
pre, cur = dummy, head
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for _ in range(m - 1):
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pre = cur
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cur = cur.next
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p, q = pre, cur
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for _ in range(n - m + 1):
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t = cur.next
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cur.next = pre
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pre = cur
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cur = t
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p.next = pre
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q.next = cur
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return dummy.next

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