LeetCode_Everyday：023 Merge k Sorted Lists

• • • 題目:
    • 示例：
    • 代碼
    • 參考
    • 此外

LeetCode Everyday：堅持價值投資，做時間的朋友！！！

題目:

合并

k

個排序連結清單，傳回合并後的排序連結清單。請分析和描述算法的複雜度。

示例：

• 示例 1:

  輸入:
  [
    1->4->5,
    1->3->4,
    2->6
  ]
  輸出: 1->1->2->3->4->4->5->6
             

代碼

方法一： 優先級隊列 題解

執行用時：88 ms, 在所有 Python3 送出中擊敗了75.77%的使用者
記憶體消耗：16.7 MB, 在所有 Python3 送出中擊敗了7.14%的使用者
           

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

from typing import List
class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        import heapq
        dummy = ListNode(0)
        p = dummy
        head = []
        for i in range(len(lists)):
            if lists[i] :
                heapq.heappush(head, (lists[i].val, i))
                lists[i] = lists[i].next
        while head:
            val, idx = heapq.heappop(head)
            p.next = ListNode(val)
            p = p.next
            if lists[idx]:
                heapq.heappush(head, (lists[idx].val, idx))
                lists[idx] = lists[idx].next
        return dummy.next

"""
For Example:    input:  [
                          1->4->5,
                          1->3->4,
                          2->6
                        ]
               output:  1->1->2->3->4->4->5->6    
"""
l1 = ListNode(1)
l1.next = ListNode(4)
l1.next.next = ListNode(5)
l2 = ListNode(1)
l2.next = ListNode(3)
l2.next.next = ListNode(4)
l3 = ListNode(2)
l3.next = ListNode(6)
lists = [l1, l2, l3]
                
solution = Solution()
result = solution.mergeKLists(lists)
print('輸出為：%d->%d->%d->%d->%d->%d->%d->%d' % \
      (result.val, result.next.val, result.next.next.val, result.next.next.next.val,\
      result.next.next.next.next.val, result.next.next.next.next.next.val,\
      result.next.next.next.next.next.next.val, result.next.next.next.next.next.next.next.val))

           

方法二： 分治 題解

執行用時：120 ms, 在所有 Python3 送出中擊敗了37.66%的使用者
記憶體消耗：23.7 MB, 在所有 Python3 送出中擊敗了7.14%的使用者
           

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

from typing import List
class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        if not lists:return 
        n = len(lists)
        return self.merge(lists, 0, n-1)
    def merge(self,lists, left, right):
        if left == right:
            return lists[left]
        mid = left + (right - left) // 2
        l1 = self.merge(lists, left, mid)
        l2 = self.merge(lists, mid+1, right)
        return self.mergeTwoLists(l1, l2)
    def mergeTwoLists(self,l1, l2):
        if not l1:return l2
        if not l2:return l1
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2

"""
For Example:    input:  [
                          1->4->5,
                          1->3->4,
                          2->6
                        ]
               output:  1->1->2->3->4->4->5->6    
"""
l1 = ListNode(1)
l1.next = ListNode(4)
l1.next.next = ListNode(5)
l2 = ListNode(1)
l2.next = ListNode(3)
l2.next.next = ListNode(4)
l3 = ListNode(2)
l3.next = ListNode(6)
lists = [l1, l2, l3]
                
solution = Solution()
result = solution.mergeKLists(lists)
print('輸出為：%d->%d->%d->%d->%d->%d->%d->%d' % \
      (result.val, result.next.val, result.next.next.val, result.next.next.next.val,\
      result.next.next.next.next.val, result.next.next.next.next.next.val,\
      result.next.next.next.next.next.next.val, result.next.next.next.next.next.next.next.val))
      
           

參考

1. https://leetcode-cn.com/problems/merge-k-sorted-lists/solution/leetcode-23-he-bing-kge-pai-xu-lian-biao-by-powcai/

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