【leetcode】链表类——合并链表

链表类——合并链表

序号	题目	难度	代码
2	Add Two Numbers	medium	python、java、c++
21	Merge Two Sorted Lists	Easy	python、java、c++
23	Merge k Sorted Lists	Hard	python、java、c++
445	Add Two Numbers II	Medium	python、java、c++

2. Add Two Numbers

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

Explanation: 342 + 465 = 807.

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

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        carry = 0
        root = n = ListNode(0)
        while l1 or l2 or carry:
            v1 = v2 = 0
            if l1:
                v1 = l1.val
                l1 = l1.next
            if l2:
                v2 = l2.val
                l2 = l2.next
            carry, val = divmod(v1+v2+carry, 10)
            n.next = ListNode(val)
            n = n.next
        return root.next       

21. Merge Two Sorted Lists

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4

Output: 1->1->2->3->4->4

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

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        if l1 and l2:
            if l1.val >l2.val:
                l1,l2 = l2,l1
            l1.next = self.mergeTwoLists(l1.next,l2)
        return l1 or l2

23. Merge k Sorted Lists

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

Example:

Input: [  1->4->5,  1->3->4,  2->6 ]

Output: 1->1->2->3->4->4->5->6

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

class Solution:
    def mergeKLists(self, lists: List[ListNode]) -> ListNode:
        pre = cur = ListNode(0)
           
        heap = []
        for i in range(len(lists)):
            if lists[i]: 
                heapq.heappush(heap, (lists[i].val, i, lists[i]))
                
        while heap:
            node = heapq.heappop(heap)
            idx = node[1]
            cur.next = node[2]  
            cur = cur.next
            
            if cur.next:
                heapq.heappush(heap, (cur.next.val, idx, cur.next))

        return pre.next

445. Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 8 -> 0 -> 7

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

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        l1 = self.reverseLL(l1)
        l2 = self.reverseLL(l2)
        car = 0
        sen = ListNode(0)
        cur = sen
        while l1 or l2 or car:
            if l1 and l2:
                tot = l1.val + l2.val + car
                l1, l2 = l1.next, l2.next
            elif l1:
                tot = l1.val + car
                l1 = l1.next
            elif l2:
                tot = l2.val + car
                l2 = l2.next
            else:
                tot = car
            cur.next, car = ListNode(tot%10), tot//10
            cur = cur.next
            
        return self.reverseLL(sen.next)
            
        
    def reverseLL(self, head):
        prev, cur = None, head
        while cur:
            cur.next, prev, cur = prev, cur, cur.next
        return prev