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
'''
'''
解题思路：
1. 归并排序的链表实现
'''
EXAMPLES = [
(
(build_list_node(1, 2, 4), build_list_node(1, 3, 4)),
build_list_node(1, 1, 2, 3, 4, 4),
),
]
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
n = ListNode(0)
temp = n
node1 = l1
node2 = l2
while not (node1 is None and node2 is None):
node1_val = node1.val if node1 else math.inf
node2_val = node2.val if node2 else math.inf
You may not modify the values in the list's node, only ondes itself may be changed.
Example:
Given 1->2->3->4, you should return the list as 2->1->4->3.
'''
'''
解题思路：
1. 交换偶数位置的节点，中间的位置存在上下关系。
2.
'''
EXAMPLES = [
(
build_list_node(1, 2, 3, 4),
build_list_node(2, 1, 4, 3),
),
]
class Solution:
def swapPairs(self, head: ListNode) -> ListNode:
if not head:
return []
node = head.next
if node is None:
r = head
else:
head.next = node.next
[
1->4->5,
1->3->4,
2->6,
]
Output: 1->1->2->3->4->5->6
'''
'''
解题思路：
1. 使用分治法将问题分解为两两合并，可以解决该题，效率可能不高。
'''
EXAMPLES = [(
[
build_list_node(1, 4, 5),
build_list_node(1, 3, 4),
build_list_node(2, 6),
],
build_list_node(1, 1, 2, 3, 4, 5, 6),
)]
class Solution:
def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
n = ListNode(0)
temp = n
node1 = l1
node2 = l2
while not (node1 is None and node2 is None):
Examples:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
OUtput: 7 -> 0 -> 8
Explanation: 342 + 465 = 807
'''
'''
总结：
1. carry_node = r.next, carry_node = ListNode(0), r.next != carry_node
'''
EXAMPLES = [
(
(build_list_node(2, 4, 3), build_list_node(5, 6, 4)),
build_list_node(7, 0, 8),
),
]
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
r = ListNode(0)
carry = 0
carry_node = r
while l1 or l2 or carry != 0:
v1 = l1.val if l1 else 0
v2 = l2.val if l2 else 0
Note:
* Only constant extra memory is allowed.
* You may not after the values in the list's nodes, only nodes itself may be
changed.
'''
'''
解题思路：
1. 属于第24题的变种，在核心判断的位置处理一下就可以了。
'''
EXAMPLES = [
(
(build_list_node(1, 2, 3, 4, 5), 2),
build_list_node(2, 1, 4, 3, 5),
),
(
(build_list_node(1, 2, 3, 4, 5), 3),
build_list_node(3, 2, 1, 4, 5),
),
]
class Solution:
def reverse(self, head):
arrays = []
node = head
while node:
Given a linked list, remove the n-th node from the end of list and return its head.
Example:
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list
becomes 1->2->3->5.
'''
'''
解题思路：
1. 执行到第 n-1 时，交换第 n 个
'''
EXAMPLES = [(build_list_node(1, 2, 3, 4, 5), )]
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
node = head
node_map = {}
i = 0
while node:
node_map[i] = node
node = node.next
i += 1
prev_node = node_map.get(i - n - 1)