leetcode 23 Merge k Sorted Lists

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本文介绍了一种高效算法用于合并K个已排序的链表为一个有序链表。通过递归地应用分治法，避免了直接顺序合并导致的时间复杂度过高的问题。

Merge k Sorted Lists

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

可以通过leetcode 21的merge Two Lists来解决。

如果从头到尾依此合并，会超时。

改用分治法来调用。链表个数l>3时，mergeTwoLists(self.mergeKLists(lists[:l/2]),self.mergeKLists(lists[l/2:l]))

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

class Solution:
    # @param two ListNodes
    # @return a ListNode
    def mergeTwoLists(self, l1, l2):
        if l1==None:
            return l2
        if l2==None:
            return l1
        if l1.val<=l2.val:
            head=l1
            l1=l1.next
        else:
            head=l2
            l2=l2.next
        p=head
        while l1 and l2:
            if l1.val<=l2.val:
                p.next=l1
                p=p.next
                l1=l1.next
            else:
                p.next=l2
                p=p.next
                l2=l2.next
        while l1:
            p.next=l1
            p=p.next
            l1=l1.next
        while l2:
            p.next=l2
            p=p.next
            l2=l2.next
        return head

    # @param a list of ListNode
    # @return a ListNode
    def mergeKLists1(self, lists):
        l=len(lists)
        if l==0:
            return None
        if l==1:
            return lists[0]
        head=self.mergeTwoLists(lists[0],lists[1])
        for i in range(2,l):
            head=self.mergeTwoLists(head,lists[i])
        return head

    # @param a list of ListNode
    # @return a ListNode
    def mergeKLists(self, lists):
        l=len(lists)
        if l==0:
            return None
        elif l==1:
            return lists[0]
        elif l==2:
            return self.mergeTwoLists(lists[0],lists[1])
        else:
            return self.mergeTwoLists(self.mergeKLists(lists[:l/2]),self.mergeKLists(lists[l/2:l]))