leetcode 【 Merge k Sorted Lists 】python 实现

题目：

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

代码：oj测试通过 Runtime: 231 ms

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    # @param a list of ListNode
    # @return a ListNode
    def mergeKLists(self, lists):
        # none
        if len(lists) == 0:
            return None
        # only one list
        if len(lists) == 1:
            return lists[0]
        # merge sort each two linked list
        l1 = self.mergeKLists(lists[:len(lists)/2])
        l2 = self.mergeKLists(lists[len(lists)/2:])
        head = self.mergeTwoLists(l1,l2)
        return head
            
    # merge two sorted linked list    
    def mergeTwoLists(self, l1, l2):
        if l1 is None:
            return l2
        if l2 is None:
            return l1
        p = ListNode(0)
        dummyhead = p
        while l1 is not None and l2 is not None:
            if l1.val < l2.val:
                p.next = l1
                l1 = l1.next
                p = p.next
            else:
                p.next = l2
                l2 = l2.next
                p = p.next
        if l1 is None:
            p.next = l2
        else:
            p.next = l1
        return dummyhead.next

思路：

总体的思路是模仿归并排序。

这里给的参数是多个链表的头结点数组，将头节点数组不断二分；直到只剩下一个头结点，返回该头节点到上一层，并在上一层中将两个有序链表合并。

将两个有序链表merge的代码和思路，在这篇日志中可以看到

http://www.cnblogs.com/xbf9xbf/p/4186905.html

另，网上还有一种用堆的方法，后续再去试探这种方法，并把另一种方法的代码补上。

另，关于该问题的算法复杂度O(nklogk) （n为单个链表的最大长度，k为链表个数），个人觉得下面这篇日志表述的比较好，共享一下

http://www.tuicool.com/articles/ZnuEVfJ