merge k sorted lists

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
import heapq
class (object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        # 这道题主要考的是heap的运用，题目不难，但是在python heap没有一个
        # 很好的实现形式，所以要在面试里熟练运用，还需要多练习
        if not lists:
            return None
        dummy = ListNode(0)
        cur = dummy
        heap = []
        # 1. add all list head node to the heap
        for index, head in enumerate(lists):
            if head:
                heapq.heappush(heap, [head.val, head])
        
        while heap:
            _, node = heapq.heappop(heap)
            if node.next:
                heapq.heappush(heap, [node.next.val, node.next])
            cur.next = node
            cur = node
        
        return dummy.next

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
import heapq
class (object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        #  还可以用divide and conquer来实现，有点像mergesort
        #  在这个mergesort的merge阶段，不是merge两个array而是merge
        #  两个list
        if not lists:
            return None
        if len(lists) == 1:
            return lists[0]
        
        mid = len(lists) / 2
        left = self.mergeKLists(lists[:mid])
        right = self.mergeKLists(lists[mid:])
        return self.merge(left, right)
        
    def merge(self, l1, l2):
        dummy = ListNode(0)
        cur = dummy
        while l1 and l2:
            if l1.val > l2.val:
                cur.next = l2
                l2 = l2.next
            else:
                cur.next = l1
                l1 = l1.next
            cur = cur.next
        if l1:
            cur.next = l1
        if l2:
            cur.next = l2
        return dummy.next