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2020.10/10.20 - 143. 重排链表.md

+给定一个单链表 `L：L0→L1→…→Ln-1→Ln` ，
+将其重新排列后变为： `L0→Ln→L1→Ln-1→L2→Ln-2→…`
+
+你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。
+
+**示例 1:**
+```
+给定链表 1->2->3->4, 重新排列为 1->4->2->3.
+```
+
+**示例 2:**
+```
+给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.
+```
+通过次数`47,851` | 提交次数`82,512`
+
+**代码实现**
+```python
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+        
+class Solution:
+    def reorderList(self, head: ListNode) -> None:
+        """
+        Do not return anything, modify head in-place instead.
+        """
+        if not head or not head.next:
+            return
+        fast,slow = head, head
+        while fast.next and fast.next.next:
+            slow = slow.next
+            fast = fast.next.next
+        p,right = slow.next,None
+        slow.next = None
+        while p:
+            right, right.next, p = p, right, p.next
+        cur = head
+        while  cur and right:
+            cur.next, right.next, cur, right = right, cur.next, cur.next, right.next
+```
+```
+来源：力扣（LeetCode）
+链接：https://leetcode-cn.com/problems/reorder-list
+著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
+```