[Leetcode][Python]23: Merge k Sorted Lists

# -*- coding: utf8 -*-‘‘‘__author__ = ‘[email protected]‘

23: Merge k Sorted Listshttps://oj.leetcode.com/problems/merge-k-sorted-lists/

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

===Comments by Dabay===

递归：每次把lists折半，分别处理左右两个子lists，然后merge。和归并排序的算法一样。时间复杂度为NlgN‘‘‘

# 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):        def merge(node1, node2):            head = node = ListNode(0)            while node1 and node2:                if node1.val < node2.val:                    node.next = node1                    node1 = node1.next                else:                    node.next = node2                    node2 = node2.next                node = node.next            if node1:                node.next = node1            elif node2:                node.next = node2            return head.next

        if len(lists) == 0:            return None        if len(lists) == 1:            return lists[0]        if len(lists) == 2:            return merge(lists[0], lists[1])

        node1 = self.mergeKLists(lists[:len(lists)/2])        node2 = self.mergeKLists(lists[len(lists)/2:])        return merge(node1, node2)

def main():    s = Solution()    ln1 = ListNode(1)    ln1.next = ListNode(10)    ln2 = ListNode(2)    ln2.next = ListNode(20)    node = s.mergeKLists([ln1, ln2])    while node:        print "%s->" % node.val,        node = node.next    print "None"

if __name__ == "__main__":    import time    start = time.clock()    main()    print "%s sec" % (time.clock() - start)