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Author: assassint2017

Merge k Sorted Lists/Merge_k_Sorted_Lists.py

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+# 第一种思路，寻找当前最小的节点，然后将一个放入新的节点内
+# 不需要额外的存储空间，时间复杂度O(N)N为所有节点的个数和
+# 4048ms 10.24%
+
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def mergeKLists(self, lists):
+        """
+        :type lists: List[ListNode]
+        :rtype: ListNode
+        """
+        # 首先剔除空链表
+        lists = [item for item in lists if isinstance(item, ListNode)]
+        if not lists:
+            return []
+        
+        temp_list = [listnode.val for listnode in lists]
+        index = temp_list.index(min(temp_list))
+        
+        mid = head = lists[index]
+        lists[index] = lists[index].next
+        if lists[index] is None:
+            lists.pop(index)
+        
+        while lists:
+            temp_list = [listnode.val for listnode in lists]
+            
+            index = temp_list.index(min(temp_list))
+
+            mid.next = lists[index]
+            mid = mid.next
+            
+            lists[index] = lists[index].next
+            if lists[index] is None:
+                lists.pop(index)
+        
+        return head
+
+# 第二种思路，根据答案的Approach 5: Merge with Divide And Conquer
+# 这里需要使用自己之前写的融合两个链表的代码
+# 96ms 66.31%
+
+class Solution:
+    def mergeKLists(self, lists):
+        """
+        :type lists: List[ListNode]
+        :rtype: ListNode
+        """
+        if not lists:
+            return lists
+        interval = 1
+        while interval < len(lists):
+            index = 0
+            while index < len(lists) - interval:
+                lists[index] = self.mergeTwoLists(lists[index], lists[index + interval])
+                index += 2 * interval
+            interval *= 2
+        return lists[0]
+
+    def mergeTwoLists(self, l1, l2):
+        if not l1:
+            return l2
+        if not l2:
+            return l1
+        main, sub = (l2, l1) if l1.val > l2.val else (l1, l2)
+        
+        main_mid = main
+        sub_mid = sub
+
+        while main_mid.next is not None:
+            if main_mid.next.val >= sub_mid.val:
+                
+                temp_node = sub_mid
+                sub_mid = sub_mid.next
+                
+                next_node = main_mid.next
+                main_mid.next = temp_node
+                temp_node.next = next_node
+                
+                main_mid = temp_node
+                
+                if sub_mid is None:
+                    return main   
+            else:
+                main_mid = main_mid.next
+        main_mid.next = sub_mid
+        return main