KaelCui的博客

剑指 Offer 56 - II. 数组中数字出现的次数 II 方法一：数学法叼炸天： c = (3(a+b+c)-(a+a+a+b+b+b+c))/2 class Solution: def singleNumber(self, nums: List[int]) -> int: return (3*sum(list(set(nums)))-sum(nums))//2 剑指 Offer 44. 数字序列中某一位的数字 链接 class Solution: def fi

剑指 Offer 18. 删除链表的节点 class Solution: def deleteNode(self, head: ListNode, val: int) -> ListNode: if head.val == val: return head.next pre, cur = head, head.next while cur and cur.val != val: pre, cur

剑指 Offer 36. 二叉搜索树与双向链表 class Solution: def treeToDoublyList(self, root: 'Node') -> 'Node': def dfs(cur): if not cur: return dfs(cur.left) # 递归左子树 if self.pre: # 修改节点引用 s

剑指 Offer 14- I. 剪绳子 class Solution: def cuttingRope(self, n: int) -> int: dp = [None, 1] for m in range (2, n + 1): j = m - 1 i = 1 max_product = 0 while i <= j: max

92. 反转链表 II Step1: The part I need to reversed is node 2 to node 4, which has n - m + 1 = 3 nodes. Therefore, I would like to maintain a window with n - m + 1 nodes with the window’s head whead and window’s tail wtail, then if whead is head, wtail would be

剑指 Offer 35. 复杂链表的复制 class Solution: def copyRandomList(self, head: 'Node') -> 'Node': if not head: return dic = {} # 3. 复制各节点，并建立 “原节点 -> 新节点” 的 Map 映射 cur = head while cur: dic[cur] = Node(cur

剑指 Offer 68 - II. 二叉树的最近公共祖先 递归解法： 模式识别：通常树的问题可以用递归解决 定义子问题：左子树和右子树都包含 p, q 最近公共祖先的定义：就是字数包含 p,q 节点 思路 当前节点已经是 p 或者 q 了，那么只要求左子树或者右子树包含另一个节点即可 左子树和右子树都包含 p 和 q 逻辑关系可以写为下图： def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode'

剑指 Offer 54. 二叉搜索树的第k大节点 # 法一：直接输出型 class Solution: def kthLargest(self, root: TreeNode, k: int) -> int: res = [] def dfs(root): if not root: return res dfs(root.left)

剑指 Offer 56 - I. 数组中数字出现的次数 class Solution: def singleNumbers(self, nums: List[int]) -> List[int]: dic = {} for num in nums: if num in dic: dic[num] += 1 else: dic[num] = 1

剑指 Offer 26. 树的子结构 class Solution: def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool: if not A or not B: return False def recur(a, b): if not b: return True

剑指 Offer 11. 旋转数组的最小数字 class Solution: def minArray(self, numbers: List[int]) -> int: i, j = 0, len(numbers) - 1 while i < j: m = (i + j) // 2 if numbers[m] > numbers[j]: i = m + 1 elif numb

剑指 Offer 10- I. 斐波那契数列 class Solution: def fib(self, n: int) -> int: a, b = 0, 1 for _ in range(n): a, b = b, a + b return int(a % 1000000007) 剑指 Offer 45. 把数组排成最小的数 class Solution: def minNumber(self, nums:

剑指 Offer 48. 最长不含重复字符的子字符串 class Solution: def lengthOfLongestSubstring(self, s: str) -> int: dic = {} res, last_match = 0, -1 for i, c in enumerate(s): if c in dic and last_match < dic[c]: last

剑指 Offer 20. 表示数值的字符串 暂不做 剑指 Offer 38. 字符串的排列 class Solution: def permutation(self, s: str) -> List[str]: self.res = [] def backtrack(s, path): if not s: self.res.append(path) seen = set()

剑指 Offer 24. 反转链表 # 双指针 # O(n)/O(1) # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def reverseList(self, head: ListNode) -> ListN