partition list medium

Author: weiy

Array/PartitionList.py

@@ -0,0 +1,71 @@
+"""
+Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.
+
+You should preserve the original relative order of the nodes in each of the two partitions.
+
+Example:
+
+Input: head = 1->4->3->2->5->2, x = 3
+Output: 1->2->2->4->3->5
+
+无脑怼。
+
+beat:
+99%.
+
+测试地址：
+https://leetcode.com/problems/partition-list/description/
+"""
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution(object):
+    def partition(self, head, x):
+        """
+        :type head: ListNode
+        :type x: int
+        :rtype: ListNode
+        """
+        
+        all_nodes = []
+        
+        while head:
+            all_nodes.append(head.val)
+            head = head.next
+        
+        less_nodes = []
+        greater_nodes = []
+        
+        for i in all_nodes:
+            if i < x:
+                less_nodes.append(i)
+            else:
+                greater_nodes.append(i)
+        
+        if less_nodes:
+            less_head = ListNode(less_nodes[0])
+        
+        head = less_head if less_nodes else None
+        
+        for i in less_nodes[1:]:
+            less_head.next = ListNode(i)
+            less_head = less_head.next
+        
+        if greater_nodes:
+            greater_head = ListNode(greater_nodes[0])
+        
+        _head = greater_head if greater_nodes else None
+        
+        for i in greater_nodes[1:]:
+            greater_head.next = ListNode(i)
+            greater_head = greater_head.next
+        
+        if head:
+            less_head.next = _head
+        
+            return head
+        return _head
+