企业官网建设流程全解析

Easy

两数之和

两数之和

class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: xdict={} for i in range(len(nums)): j=target-nums[i] if j in xdict.keys(): return [i,xdict[j]] else: xdict[nums[i]]=i

有效的括号

有效的括号

class Solution: def isValid(self, s: str) -> bool: map_dict={ ')':'(', '}':'{', ']':'[', } st=[] for item in s: if item not in map_dict.keys(): st.append(item) else: if not st: return False pop_item=st.pop() if pop_item!=map_dict[item]: # 如果是嵌套括号的话，不能交错嵌套。如果不是交错嵌套的，内层的会正确被 pop 出来，所以只需要 pop 最后的那个就可以了。 return False return not st

合并两个有序链表

合并两个有序链表

# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]: # if not list1 and not list2: # return None curr=ListNode() head=curr while list1 and list2: if list1.val<list2.val: curr.next=list1 list1=list1.next else: curr.next=list2 list2=list2.next curr=curr.next if list1: curr.next=list1 if list2: curr.next=list2 return head.next

爬楼梯

爬楼梯

class Solution: def climbStairs(self, n: int) -> int: if n==1: return 1 if n==2: return 2 dp=[0]*n dp[0]=1 dp[1]=2 for i in range(2,n): dp[i]=dp[i-1]+dp[i-2] return dp[n-1]

二叉树的中序遍历

二叉树的中序遍历

# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]: ret=[] def dfs(node): if not node: return dfs(node.left) ret.append(node.val) dfs(node.right) dfs(root) return ret

对称二叉树

对称二叉树

# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def isSymmetric(self, root: Optional[TreeNode]) -> bool: if not root: return True def search(t1,t2): if not t1 and not t2: return True if not t1 or not t2: return False return t1.val==t2.val and search(t1.left,t2.right) and search(t1.right,t2.left) return search(root.left,root.right)

二叉树的最大深度

二叉树的最大深度

# Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def maxDepth(self, root: Optional[TreeNode]) -> int: if not root: return 0 left_dep=self.maxDepth(root.left) right_dep=self.maxDepth(root.right) return max(left_dep,right_dep)+1 # 加一的根本原因是：为了把当前节点所在的这一层计算进去。

买卖股票的最佳时机

买卖股票的最佳时机

class Solution: def maxProfit(self, prices: List[int]) -> int: max_ret=0 min_cost=float('inf') n=len(prices) for i in range(1,n): min_cost=min(min_cost,prices[i-1]) max_ret=max(max_ret,prices[i]-min_cost) return max_ret