83

Author: tech-cow

README.md

@@ -34,3 +34,4 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 |  #  | Title | Solution | Time | Space | Difficulty |Tag| Note|
 |-----|-------| -------- | ---- | ------|------------|---|-----|
 |83|[Remove Duplicates from Sorted List](https://leetcode.com/problems/remove-duplicates-from-sorted-list/#/description)| [Python](./linkedlist/deleteDuplicates.py) | _O(n)_| _O(1)_  | Easy | ||
+|148|[Sort List](https://leetcode.com/problems/sort-list/#/description)| [Python](./linkedlist/sortList.py) | _O(nlogn)_| _O(1)_  | Medium | ||

linkedlist/sortList.py

@@ -0,0 +1,86 @@
+# Author: Yu Zhou
+# 148. Sort List
+
+# Sort a linked list in O(n log n) time using constant space complexity.
+
+# 思路
+# 用链表的方式切分，然后递归Split，之后Merge
+# 这道题要注意是切分的时候，要写个Prev的函数用来储存Slow的原点
+
+# Time: O(nlogn)
+# Space: O(1)
+
+# ****************
+# Final Solution *
+# ****************
+class Solution(object):
+    def sortList(self, head):
+        if not head or not head.next:
+            return head
+
+        prev = None
+        fast = slow = head
+        while fast and fast.next:
+            prev = slow
+            slow = slow.next
+            fast = fast.next.next
+
+        mid = slow
+        prev.next = None #截断
+
+        l = self.sortList(head) #递归
+        r = self.sortList(mid)
+        return self.merge(l, r)
+
+
+    #这是一种类似于创建一个头的DummyNode，把Head设置成Prev，Cur设置成head.next
+    def merge(self, l, r):
+        guard = cur = ListNode(None)
+        while l and r:
+            if l.val <= r.val:
+                cur.next = l
+                l = l.next
+            else:
+                cur.next = r
+                r = r.next
+            cur = cur.next
+        cur.next = l or r
+        return guard.next
+
+        vhead = curr = ListNode(0)
+        while l1 and l2:
+            if l1.val <= l2.val:
+                curr.next = l1
+                l1 = l1.next
+            else:
+                curr.next = l2
+                l2 = l2.next
+            curr = curr.next
+        curr.next = l1 or l2
+        return vhead.next
+
+
+# ***************************************
+# The following code is an fail attempt *
+# ***************************************
+class Solution(object):
+    def sortList(self, head):
+        if not head or not head.next:
+            return head
+
+        fast = slow = head
+        #这里错在没有设置一个prev来保存Slow的值
+        #每次while结束后，slow就会变成slow.next
+        #假如想在slow的地方切断，就必须设置一个Prev
+        #来保存slow的值，下方 mid = slow.next
+        #其实已经是在经历过while 后的slow.next的next了
+        while fast and fast.next:
+            slow = slow.next
+            fast = fast.next.next
+
+        mid = slow.next
+        slow.next = None #截断
+
+        l = self.sortList(head) #递归
+        r = self.sortList(mid)
+        return self.merge(l, r)