# 矩阵简单题

770. Maximum and Minimum

Given a matrix, return the maximum number and the minimum number.

Given a matrix:

return

ifmatrix == []:

return[]

nums = [,]

nums[] = matrix[][]

nums[1] = matrix[][]

length =len(matrix)

foriinrange(length):

nums[]=max(nums[],max(matrix[i]))

nums[1]=min(nums[1],min(matrix[i]))

returnnums

737. 查找矩阵

result = Matrix[]

length =len(Matrix)

foriinrange(1,length):

forjinresult:

ifjnot inMatrix[i]:

result.remove(j)

returnresult[]

importos

fromseleniumimportwebdriver

driver = webdriver.Chrome()

driver.maximize_window()

driver.get("https://www.zhihu.com/signup?next=%2F")

driver.find_element_by_xpath('//*[@id="root"]/div/main/div/div/div/div[2]/div[2]/span').click()

driver.find_element_by_xpath('//*[@id="root"]/div/main/div/div/div/div[2]/div[1]/form/button').click()

print(driver.title)

os.system("pause")

• 发表于:
• 原文链接http://kuaibao.qq.com/s/20180321G1RIHK00?refer=cp_1026
• 腾讯「云+社区」是腾讯内容开放平台帐号（企鹅号）传播渠道之一，根据《腾讯内容开放平台服务协议》转载发布内容。
• 如有侵权，请联系 yunjia_community@tencent.com 删除。

2022-05-22

2022-05-22

2018-03-12

2022-05-22

2022-05-22

2022-05-22

2022-05-22

2022-05-22

2022-05-22

2022-05-22

2022-05-22