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feat: add python and java solutions to leetcode problem: No.0143
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‎README.md

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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. [重排链表](/solution/0100-0199/0143.Reorder%20List/README.md)
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1. [回文链表](/solution/0200-0299/0234.Palindrome%20Linked%20List/README.md)
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1. [环形链表](/solution/0100-0199/0141.Linked%20List%20Cycle/README.md)
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1. [环形链表 II](/solution/0100-0199/0142.Linked%20List%20Cycle%20II/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. [Reorder List](/solution/0100-0199/0143.Reorder%20List/README_EN.md)
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1. [Palindrome Linked List](/solution/0200-0299/0234.Palindrome%20Linked%20List/README_EN.md)
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1. [Linked List Cycle](/solution/0100-0199/0141.Linked%20List%20Cycle/README_EN.md)
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1. [Linked List Cycle II](/solution/0100-0199/0142.Linked%20List%20Cycle%20II/README_EN.md)

‎solution/0100-0199/0143.Reorder List/README.md

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<!-- 这里可写通用的实现逻辑 -->
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先通过快慢指针找到链表中点，将链表划分为左右两部分。之后反转右半部分的链表，然后将左右两个链接依次连接即可。
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<!-- tabs:start -->
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### **Python3**
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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, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def reorderList(self, head: ListNode) -> None:
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"""
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Do not return anything, modify head in-place instead.
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"""
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if head is None or head.next is None:
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return
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slow, fast = head, head.next
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# 快慢指针找到链表中点
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while fast and fast.next:
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slow, fast = slow.next, fast.next.next
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cur = slow.next
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slow.next = None
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pre = None
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# cur 指向右半部分的链表，反转
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while cur:
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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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cur = head
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# 将左右链表依次连接
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while pre:
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t1 = cur.next
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cur.next = pre
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cur = t1
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t2 = pre.next
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pre.next = t1
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pre = t2
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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() {}
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* ListNode(int val) { this.val = val; }
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* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
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* }
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*/
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class Solution {
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public void reorderList(ListNode head) {
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if (head == null || head.next == null) {
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return;
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}
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ListNode slow = head;
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ListNode fast = head.next;
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while (fast != null && fast.next != null) {
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slow = slow.next;
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fast = fast.next.next;
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}
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ListNode cur = slow.next;
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slow.next = null;
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ListNode pre = null;
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while (cur != null) {
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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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cur = head;
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while (pre != null) {
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ListNode t1 = cur.next;
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cur.next = pre;
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cur = t1;
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ListNode t2 = pre.next;
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pre.next = cur;
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pre = t2;
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}
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}
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}
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```
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### **...**

‎solution/0100-0199/0143.Reorder List/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -31,13 +31,83 @@ Given 1-&gt;2-&gt;3-&gt;4-&gt;5, reorder it to 1-&gt;5-&gt;2-&gt;4-&gt;3.
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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, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def reorderList(self, head: ListNode) -> None:
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"""
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Do not return anything, modify head in-place instead.
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"""
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if head is None or head.next is None:
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return
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slow, fast = head, head.next
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while fast and fast.next:
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slow, fast = slow.next, fast.next.next
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cur = slow.next
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slow.next = None
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pre = None
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while cur:
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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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cur = head
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while pre:
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t1 = cur.next
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cur.next = pre
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cur = t1
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t2 = pre.next
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pre.next = t1
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pre = t2
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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() {}
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* ListNode(int val) { this.val = val; }
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* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
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* }
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*/
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class Solution {
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public void reorderList(ListNode head) {
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if (head == null || head.next == null) {
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return;
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}
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ListNode slow = head;
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ListNode fast = head.next;
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while (fast != null && fast.next != null) {
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slow = slow.next;
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fast = fast.next.next;
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}
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ListNode cur = slow.next;
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slow.next = null;
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ListNode pre = null;
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while (cur != null) {
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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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cur = head;
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while (pre != null) {
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ListNode t1 = cur.next;
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cur.next = pre;
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cur = t1;
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ListNode t2 = pre.next;
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pre.next = cur;
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pre = t2;
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}
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}
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}
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```
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### **...**

‎solution/0100-0199/0143.Reorder List/Solution.java

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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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* ListNode() {}
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* ListNode(int val) { this.val = val; }
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* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
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* }
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*/
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class Solution {
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public void reorderList(ListNode head) {
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if (head == null || head.next == null || head.next.next == null) {
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if (head == null || head.next == null) {
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return;
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}
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ListNode dummy = new ListNode(-1);
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dummy.next = head;
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ListNode slow = head;
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ListNode fast = head;
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ListNode pre = dummy;
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ListNode fast = head.next;
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while (fast != null && fast.next != null) {
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fast = fast.next.next;
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slow = slow.next;
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pre = pre.next;
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}
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pre.next = null;
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// 将后半段的链表进行 reverse
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ListNode rightPre = new ListNode(-1);
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rightPre.next = null;
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ListNode t = null;
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while (slow != null) {
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t = slow.next;
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slow.next = rightPre.next;
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rightPre.next = slow;
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slow = t;
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fast = fast.next.next;
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}
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ListNode p1 = dummy.next;
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ListNode p2 = p1.next;
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ListNode p3 = rightPre.next;
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ListNode p4 = p3.next;
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while (p1 != null) {
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p3.next = p1.next;
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p1.next = p3;
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if (p2 == null) {
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break;
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}
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p1 = p2;
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p2 = p1.next;
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p3 = p4;
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p4 = p3.next;
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ListNode cur = slow.next;
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slow.next = null;
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ListNode pre = null;
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while (cur != null) {
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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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if (p4 != null) {
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p1.next.next = p4;
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cur = head;
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while (pre != null) {
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ListNode t1 = cur.next;
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cur.next = pre;
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cur = t1;
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ListNode t2 = pre.next;
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pre.next = cur;
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pre = t2;
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}
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head = dummy.next;
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}
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}

‎solution/0100-0199/0143.Reorder List/Solution.py

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@@ -0,0 +1,31 @@
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def reorderList(self, head: ListNode) -> None:
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"""
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Do not return anything, modify head in-place instead.
10+
"""
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if head is None or head.next is None:
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return
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slow, fast = head, head.next
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while fast and fast.next:
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slow, fast = slow.next, fast.next.next
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cur = slow.next
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slow.next = None
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pre = None
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while cur:
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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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cur = head
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while pre:
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t1 = cur.next
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cur.next = pre
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cur = t1
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t2 = pre.next
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pre.next = t1
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pre = t2

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