Question

def listmerge(L1, L2):
    '''(CustomerNode, CustomerNode) -> CustomerNode
    Merge the linked lists headed by L1 and L2 into a single list with ticket_num in
    increasing order. Return the head of the merged list.
    REQ: Lists headed by L1 and L2 are sorted by ticket_num, all ticket_num    values are unique.
    '''

    current = L1
    while current.next != None:
        current = current.next
    current.next = L2
    return L1

客户节点的一个例子就像LinkedList，除了像这样：

list1 = CustomerNode(priority, data, next = None)

基本上我的排序优先级。最低的数字应该是头部。

到目前为止，我刚刚进行了合并。

Answer 1

我在想，在合并节点之前，您可以先按照要排序的类别进行排序，然后使用listmerge(L1, L2)函数进行合并。

假设我们有这些节点：

class Node():
    def __init__(self, name, data, next=None):
        self.name = name
        self.data = data
        self.next = next

注意：我已将priority替换为name

现在我要创建一个列表来保存所有这些节点而不进行任何排序：

nodeList = [<__main__.Node object at 0x1021e7320>, <__main__.Node object at 0x1021e7278>, <__main__.Node object at 0x1021e72b0>, <__main__.Node object at 0x1021e7240>, <__main__.Node object at 0x1021e72e8>]

然后我可以按照我要排序的字段进行排序。我将假设按data排序：

# Helper function
def getData(node):
    return node.data

sortedList = sorted(nodeList, key=getData)

最后通过迭代sortedList并将每个节点传递到listmerge函数来进行合并。

显示上述结果的示例：

# This is the list full of unordered objects. I want to sort by the data integer values
mynodelist = [Node('a', 10), Node('b', 5), Node('c', 7), Node('d', 20), Node('e', 2)]


sortedList = sorted(mynodelist, key=getData)

for o in sortedList:
    print(o.data)

### OUTPUT ###
# 2
# 5
# 7
# 10
# 20

我如何排序链表？

1 个答案: