LeetCode----Merge k Sorted Lists

Merge k Sorted Lists

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

分析1：

利用上一道题（合并两个有序链表）代码，进行两两合并，最后得到一个链表。代码如下，然而一直超时。

超时的代码：

    def mergeKLists(self, lists):        """        :type lists: List[ListNode]        :rtype: ListNode        """        if len(lists) == 0:            return None        else:            l3 = None            for l in lists:                l3 = reduce(self.mergetTwoLists, lists)        return l3.next

分析2:

利用最小堆的特性，将每个链表放入最小堆中，每次从堆中取出一个ListNode p，同时往里面插入p的后序ListNode（如果存在的话）。我按算法导论的堆的伪码，用Python实现了一个最小堆完成了这道题。如果你看下面这段代码的话，你可以使用Python的heapq模块中的函数替换我写的渣渣代码，同时可以去看看heapq中的实现。

代码：

def Parent(i):    return i/2def Left(i):    return 2*idef Right(i):    return 2*i + 1def minHeapify(h, i):    l = Left(i)    r = Right(i)    if l < len(h) and h[l] < h[i]:        minlst = l    else:        minlst = i    if r < len(h) and h[r] < h[minlst]:        minlst = r    if minlst != i:        h[i], h[minlst] = h[minlst], h[i]        minHeapify(h, minlst)def buildMinHeap(h):    half_len = len(h) / 2    while half_len >= 0:        minHeapify(h, half_len)        half_len -= 1    return hdef upAdjust(h, i):    while i >= 0 and h[Parent(i)] > h[i]:        h[Parent(i)], h[i] = h[i], h[Parent(i)]        i = Parent(i)def heapPush(h, item):    h.append(item)    upAdjust(h, len(h) - 1)def heapPop(h):    if len(h) == 0:        return None    min_val = h[0]    h[0] = h[len(h) - 1]    del h[len(h) - 1]    minHeapify(h, 0)    return min_val# Definition for singly-linked list.class ListNode(object):    def __init__(self, x):        self.val = x        self.next = Noneclass Solution(object):    def mergeKLists(self, lists):        """        :type lists: List[ListNode]        :rtype: ListNode        """        # from heapq import heappush, heappop, heapify        h = [(l.val, l) for l in lists if l]        buildMinHeap(h)        head = rl = ListNode(0)        while h:            rl.next = heapPop(h)[1]            print rl.next.val            rl = rl.next            if rl.next:                heapPush(h, (rl.next.val, rl.next))        return head.next