83

Author: tech-cow

README.md

@@ -28,3 +28,9 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 |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/reorder.py) | _O(n)_| _O(1)_  | Medium | ||
 |24|[Swap Nodes in Pairs](https://leetcode.com/problems/swap-nodes-in-pairs/#/solutions)| [Python](./linkedlist/swapPairs.py) | _O(n)_| _O(1)_  | Medium | |[Note](http://imgur.com/a/4G6ng)|
+
+
+## 4/15 Tasks (LinkedList Medium)
+|  #  | 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 | ||

linkedlist/deleteDuplicates.py

@@ -0,0 +1,51 @@
+# Author: Yu Zhou
+# 83. Remove Duplicates from Sorted List
+
+# Given a sorted linked list, delete all duplicates such that each element appear only once.
+#
+# For example,
+# Given 1->1->2, return 1->2.
+# Given 1->1->2->3->3, return 1->2->3.
+
+# 思路，不需要删除这个Node，只需要将Pointer指向之后的一个Node就行
+# 中间有个容易出错的就是，不要每次While循环，自动位移当前的Pointer
+# 在比对完当前Node和后面的所有Node直到没有重复的时候
+# 再 cur = cur.next
+
+# ****************
+# Final Solution *
+# ****************
+class Solution(object):
+    def deleteDuplicates(self, head):
+        """
+        :type head: ListNode
+        :rtype: ListNode
+        """
+        cur = head
+        while cur and cur.next:
+            if cur.val == cur.next.val:
+                cur.next = cur.next.next
+            else:
+                cur = cur.next
+        return head
+
+# ***************************************
+# The following code is an fail attempt *
+# ***************************************
+class Solution(object):
+    def deleteDuplicates(self, head):
+        """
+        :type head: ListNode
+        :rtype: ListNode
+        """
+        cur = head
+
+        while cur and cur.next:
+            if cur.val == cur.next.val:
+                cur.next = cur.next.next
+        #问题出在这里
+        #cur不应该在每次比对以后直接位移，应该等没有
+        #duplicate以后再位移，所以这里要加Else
+        #不加的后果是，底下之中corner case会失败: [1,1,1]
+            cur = cur.next
+        return head