swap nodes update

Author: tech-cow

README.md

@@ -21,9 +21,10 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 |2|[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/#/description) | [Python](./linkedlist/addTwoNumbers.py) | _O(n)_| _O(n)_  | Medium |CC189| |
 |445| [Add Two Numbers II](https://leetcode.com/problems/add-two-numbers-ii/#/description)| [Python](./linkedlist/addTwoNumbersTwo.py) | _O(n)_| _O(n)_  | Medium |CC189 |Stack|
 
-## Daily Task
+## 4/14 Tasks (LinkedList Medium)
 |  #  | Title | Solution | Time | Space | Difficulty |Tag| Note|
 |-----|-------| -------- | ---- | ------|------------|---|-----|
 |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 | ||
+|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 | ||

linkedlist/swapPairs.py

@@ -0,0 +1,77 @@
+# 24. Swap Nodes in Pairs
+# Author: Yu Zhou
+
+# Given a linked list, swap every two adjacent nodes and return its head.
+#
+# For example,
+# Given 1->2->3->4, you should return the list as 2->1->4->3.
+#
+# 思路
+
+# 这题的关键是处理每个Node的指针关系
+# 设置3个pointer， p0也就是prev，p1是head, p2是next
+# 然后把关系通过纸画下来，就能够理解个个pointer的关系了。
+# 之前我的想法特别奇葩，在下面的代码里，感觉也没写对。。
+
+# ****************
+# Final Solution *
+# ****************
+class Solution(object):
+    def swapPairs(self, head):
+        if not head:
+            return head
+
+        guard = p0 = ListNode(0)
+        guard.next = head
+
+        while p0.next and p0.next.next:
+            p1 = p0.next  #第一个交换数
+            p2 = p0.next.next  #第2个交换数
+
+            p0.next = p2
+            p1.next = p2.next
+            p2.next = p1
+
+            p0 = p0.next.next
+
+        return guard.next
+
+# ***************************************
+# The following code is an fail attempt *
+# ***************************************
+class Solution(object):
+    def swapPairs(self, head):
+        if not head:
+            return head
+
+        dummy1 , dummy2 = ListNode(0), ListNode(0)
+
+        cur = head.next
+        while cur:
+            dummy1.next = cur
+            if cur.next:
+                cur = cur.next.next
+
+        cur = head
+        while cur and cur.next:
+            dumm2.next = cur
+            cur = cur.next.next
+
+        #merge
+        dummy = ListNode(0)
+        dummy.next = head
+        while dummy1 and dummy2:
+            dummy.next = dummy1
+            dummy = dummy.next
+            dummy1 = dummy1.next
+
+            dummy.next = dummy2
+            dummy = dummy.next
+            dummy2 = dummy2.next
+
+        if dummy1:
+            dummy.next = dummy1
+        if dummy2:
+            dummy.next = dummy2
+
+        return dummy.next