[21. 合併兩個有序鍊錶](
[2. 兩數相加](
[61. 旋轉鍊錶](
可見，如果經常需要新增或刪除結點，鍊錶更好，經常按索引訪問元素，陣列更好
class
solution
:def
mergetwolists
(self, l1: listnode, l2: listnode)
-> listnode:
# 設定合併後的表頭
new_head = listnode(-1
) head = new_head
# l1 和 l2總有乙個會先走到頭
while l1 and l2:
if l1.val < l2.val:
head.
next
= l1
l1 = l1.
next
else
: head.
next
= l2
l2 = l2.
next
head = head.
next
# 此時只有一條鍊錶走完，剩下的鍊錶直接拼接在最後面即可
if l1:
head.
next
= l1
else
: head.
next
= l2
return new_head.
next
class
solution
:def
addtwonumbers
(self, l1: listnode, l2: listnode)
-> listnode:
# 用0補齊短鍊表
p1 = l1
p2 = l2
prev_p1 =
none
prev_p2 =
none
while p1 or p2:
ifnot p1 and p2:
prev_p1.
next
= listnode(0)
p1 = prev_p1.
next
ifnot p2 and p1:
prev_p2.
next
= listnode(0)
p2 = prev_p2.
next
if p1 and p2:
prev_p1 = p1
prev_p2 = p2
p1 = p1.
next
p2 = p2.
next
# 齊位運算
p1 = l1
p2 = l2
carry =
0 prev_p1 =
none
while p1 and p2:
ifcarry =
(p1.val + p2.val + carry)
>=
10 p1.val =
(p1.val + p2.val + carry)%10
if ifcarry:
carry =
1else
: carry =
0 prev_p1 = p1
p1 = p1.
next
p2 = p2.
next
ifnot p1 and carry ==1:
prev_p1.
next
= listnode(1)
return l1
# definition for singly-linked list.
# class listnode:
# def __init__(self, val=0, next=none):
# self.val = val
# self.next = next
class
solution
:def
rotateright
(self, head: listnode, k:
int)
-> listnode:
ifnot head or
not head.
next
:return head
# 確定長度
p1 = head
length =
0while p1:
length +=
1 p1 = p1.
next
pre_tail =
none
tail = head
for _ in
range
(k%length)
:while tail.
next
: pre_tail = tail
tail = tail.
next
pre_tail.
next
=none
# 尾結點新增進頭結點
tail.
next
= head
head = tail
return head
Leetcode刷題鍊錶之環形鍊錶
給定乙個鍊錶，判斷鍊錶中是否有環。定義兩個指標，從頭節點開始，兩個指標都向右移動，但是設定他們的移動速度不一樣，如果為環形鍊錶，則指標肯定會相遇。若為直鏈表，兩個指標至少有乙個為空。definition for singly linked list.class listnode public cla...
Leetcode鍊錶刷題之 206 反轉鍊錶
反轉乙個單鏈表。示例 輸入 1 2 3 4 5 null 輸出 5 4 3 2 1 null 高階 你可以迭代或遞迴地反轉鍊錶。你能否用兩種方法解決這道題？解題思路 原地旋轉鍊錶，可以構建兩個指標，乙個指向當前結點的前乙個，乙個指向當前的結點，每次迴圈的時候，找到當前鍊錶的下乙個，在翻轉的過程中，將...
LeetCode刷題筆記 328 奇偶鍊錶
給定乙個單鏈表，把所有的奇數節點和偶數節點分別排在一起。請注意，這裡的奇數節點和偶數節點指的是節點編號的奇偶性，而不是節點的值的奇偶性。請嘗試使用原地演算法完成。你的演算法的空間複雜度應為 o 1 時間複雜度應為 o nodes nodes 為節點總數。示例 輸入 2 1 3 5 6 4 7 nul...