这题有毒

Author: tech-cow

README.md

@@ -26,3 +26,4 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 |-----|-------| -------- | ---- | ------|------------|---|-----|
 |142|[Linked List Cycle II](https://leetcode.com/problems/linked-list-cycle-ii/#/description)| [Python](./linkedlist/detectCycleii.py) | _O(n)_| _O(1)_  | Medium |CC189 |[Video](https://www.youtube.com/watch?v=iZVBVCpmugI&t=1s)|
 |328|[Odd Even Linked List](https://leetcode.com/problems/odd-even-linked-list/#/description)| [Python](./linkedlist/oddEvenList.py) | _O(n)_| _O(1)_  | Medium | ||
+|143|[Reorder List](https://leetcode.com/problems/reorder-list/#/description)| [Python](./linkedlist/reorderList.py) | _O(n)_| _O(1)_  | Medium | ||

linkedlist/reorderList.py

@@ -0,0 +1,99 @@
+# 143. Reorder List
+# Author: Yu Zhou
+
+# Given a singly linked list L: L0→L1→…→Ln-1→Ln,
+# reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…
+#
+# You must do this in-place without altering the nodes' values.
+#
+# For example,
+# Given {1,2,3,4}, reorder it to {1,4,2,3}.
+
+# 思路
+# 这题的思路是将Node分制
+# 设置俩Runner，取中间值，在对中间值后续的Node进行reverse，之后根据顺序合并
+# 切记在将一个链表切分的时候，一定要将左边的链表末端指向Null，否则就没切断。
+
+# ****************
+# Final Solution *
+# ****************
+class Solution(object):
+    def reorderList(self, head):
+        if not head:
+            return
+        mid = self.findMid(head)
+        last_node = self.reverse(mid)
+        self.merge(head, last_node)
+
+    def findMid(self,head):
+        slow = fast = head
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+        mid = slow.next
+        slow.next = None
+
+        return mid
+
+    def reverse(self, node):
+        prev = None
+        cur = node
+
+        while cur:
+            next = cur.next
+            cur.next = prev
+            prev = cur
+            cur = next
+        return prev
+
+    # Merge应该设置Next的Guardnode，因为每次我们将Cur Node指向两一个链表的Head Node的时候
+    # 我们就缺失原有链表的Next的指针。
+    def merge(self, head, end):
+        cur1, cur2 = head, end
+        while cur2:
+            temp1, temp2 = cur1.next, cur2.next
+            cur1.next = cur2
+            cur2.next = temp1
+            cur1 = temp1
+            cur2 = temp2
+        return
+
+# ***************************************
+# The following code is an fail attempt *
+# ***************************************
+class Solution(object):
+    def reorderList(self, head):
+        if not head:
+            return
+        mid = self.mid_finder(head)
+        last_node = self.reverse(mid)
+        self.merge(head, last_node)
+
+    def mid_finder(self,head):
+        slow = fast = head
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+            #这里犯错了，应该把左半边的链表指向Null，不然切不断
+            # 缺了slow.next = None
+        return slow
+
+    def reverse(self, node):
+        prev = None
+        cur = node
+        while cur:
+            next = cur.next
+            cur.next = prev
+            prev = cur
+            cur = next
+        return prev
+
+    def merge(self, head, last):
+
+        first = head
+        while last:
+           first.next = last
+           first = first.next
+           last.next = first
+           last = last.next
+        return