leetcode-hard-ListNode-23. Merge k Sorted Lists

mycode   91.2%

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

class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        
        res = []
        for li in lists:
            while li:
                res.append(li.val)
                li = li.next
        if not res :
            return None
        res = sorted(res)
        print(res)
        length = len(res)
        i = 0
        dummy = head = ListNode(-1)
        while i < length: 
            head.next = ListNode(res[i])
            head = head.next   
            i += 1
        return dummy.next

但是下面的这个代码最后的链表就只有1一个val

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

class Solution(object):
    def mergeKLists(self, lists):
        """
        :type lists: List[ListNode]
        :rtype: ListNode
        """
        
        res = []
        for li in lists:
            while li:
                res.append(li.val)
                li = li.next
        if not res :
            return None
        res = sorted(res)
        print(res)
        length = len(res)
        i = 0
        dummy = head = ListNode(res[0])
        head = head.next
        while i < length - 1:
            i += 1
            #print(i)
            head = ListNode(res[i])
            #print(head.val)
            head = head.next
        return dummy

Input:[[1,4,5],[1,3,4],[2,6]]

Output:[1]

Expected:[1,1,2,3,4,4,5,6]

Stdout:[1, 1, 2, 3, 4, 4, 5, 6]

 

 

参考

class Solution:
    def mergeKLists(self, lists):
        nums, dum = [], ListNode(0)
        p = dum
        for l in lists:
            while l:
                nums.append(l)
                l = l.next
        for i in sorted(nums, key = lambda l: l.val):
            p.next = i
            p = p.next
        return dum.next