HuWF的博客

牛客网链接# -*- coding:utf-8 -*-#时间复杂度偏高，但可通过class Solution: # array 二维列表 def Find(self, target, array): # write code here for row in array: if target in row : retu...

牛客网链接# -*- coding:utf-8 -*-# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: # 返回二维列表[[1,2],[4,5]] ...

牛客网链接# -*- coding:utf-8 -*-# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def isSymmetrical(self, ...

牛客网链接# -*- coding:utf-8 -*-# class TreeLinkNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = None# self.next = Noneclass Solutio...

牛客网链接# -*- coding:utf-8 -*-# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def TreeDepth(self, pRoo...

牛客网链接# -*- coding:utf-8 -*-# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: # 返回从上到下每个节点值列表，例：[1,2,3...

牛客网题目链接# -*- coding:utf-8 -*-# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: # 返回镜像树的根节点 def Mi...

题目链接：https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=13&tqId=11157&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking# -*- ...

# -*- coding:utf-8 -*-# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def HasSubtree(self, pRoot1, pR...

牛客网链接# -*- coding:utf-8 -*-class Solution: def Power(self, base, exponent): # write code here #return base**exponent if base==0:return 0 ans=1 for i in ...

牛客网链接思路# -*- coding:utf-8 -*-class Solution: def NumberOf1(self, n): # write code here if n>=0 : return bin(n).count('1') else: return bin...

牛客网链接思路：对于相似的斐波那契问题（青蛙跳台阶、变态青蛙跳台阶、铺砖问题），都可以先给出前几个初始值，然后按照斐波那契规律递推即可。# -*- coding:utf-8 -*-class Solution: def rectCover(self, number): #对于相似的斐波那契亚问题，都可以先给出前几个初始值，然后按照规律递推即可 #...

牛客网链接思路： 因为n级台阶，第一步有n种跳法：跳1级、跳2级、到跳n级 跳1级，剩下n-1级，则剩下跳法是f(n-1) 跳2级，剩下n-2级，则剩下跳法是f(n-2) 所以f(n)=f(n-1)+f(n-2)+...+f(1) 因为f(n-1)=f(n-2)+f(n-3)+...+f(1) 所...

牛客网链接思路：手动计算给出前几个初始值，然后按照斐波那契规律递推即可。注意，对于不同的具体斐波那契类问题问题，其前几个初始值可能不同！# -*- coding:utf-8 -*-class Solution: def jumpFloor(self, number): # write code here fibs=[0,1,2] ...

牛客网链接# -*- coding:utf-8 -*-class Solution: def Fibonacci(self, n): # write code here #若直接使用递归，则导致计算时间溢出,因此用循环的方法实现递归 fibs=[0,1] while n>=len(fibs): ...

牛客网链接# -*- coding:utf-8 -*-class Solution: def minNumberInRotateArray(self, rotateArray): # write code here if rotateArray==[] : return [] return sorted(rotateArray)[0]...

牛客网链接# -*- coding:utf-8 -*-# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def IsBalanced_Solution(...