4.23 10:00更新。编程题1的Python实现。仅供參考。源代码见页尾

4.23 20:35更新，编程题2的Python实现。源代码见尾页

百度的题还是很偏重算法的。总体来讲难度比較高。尤其是编程题，以下附上原题：

选择题

问答题

主观题

编程题

编程题1源代码

#coding:utf-8

data = http://www.mamicode.com/[]>
?)'], [1,2], [2,8]]
    data.append(item) #data = http://www.mamicode.com/[  ['(?
?)', [1,2], [2,8]], ......  ]

# 生成全部括号的可能性
def allThePosibility(theString,data):
    # 对于该问号有改成)和(这两种可能
    if theString.count('?
') == 0:
        data.append(theString)
    else:
        theStringToLeft = ''
        theStringToRight = ''
        theIndex = theString.index('?') # 第一个问号的位置
        theStringToLeft = theString[:theIndex] + '(' + theString[theIndex+1:]
        #print theStringToLeft
        theStringToRight = theString[:theIndex] + ')' + theString[theIndex + 1:]
        #print theStringToRight
        allThePosibility(theStringToLeft,data)
        allThePosibility(theStringToRight,data)
        return data # ['((()', '(())', '()()', '()))']

# 是否正则化
def isRegularization(theString): # theString = '((()'
    if theString.count('(') != theString.count(')'):
        return 0
    stack = []  # 设置一个栈
    for alphabet in theString:
        if alphabet == ')' and stack == []:
            return 0
        else:
            if alphabet == '(':
                stack.append(0) # 入栈
            else: # 遇到右括号
                stack.pop() # 出栈
    if stack != []:
        return 0
    else:
        return theString

# 每一个问号的位置
def positionOfQuestionMark(theString): # theString = '(?
?
)'
    i = 0
    position = []
    while True:
        if '?
' in theString[i:]:
            theIndex = theString[i:].index('?
') # 更新下一个问号的位置
            i += theIndex
            position.append(i)
            i += 1
        else:
            break
    return position # [1,2]

# 处理数据
for item in data: # item = ['(??)', [1,2], [2,8]]

    regularzations = []

    # 全部括号的位置
    position = positionOfQuestionMark(item[0]) # position = [1,2]
    # 列出全部能加括号的情况
    posibilities = allThePosibility(item[0], data=http://www.mamicode.com/[]) # posibilities = ['((()', '(())', '()()', '()))']>


编程题2
#coding:utf-8

# 百度笔试题2

# 先处理输入
theString = raw_input()
base = int(theString.split(' ')[0]) # string型
luckyNum = int(theString.split(' ')[1]) # string型

if base < luckyNum:
    print luckyNum
elif base > luckyNum and len(str(base)) > len(str(luckyNum)):
    x = len(str(luckyNum)) # 幸运数的位数
    if int(str(base)[len(str(base)) - x : ]) <= luckyNum: # 假设base的后x位小于等于幸运数
        print str(base)[:len(str(base)) - x] + str(luckyNum)
    else: # base的后x为大于幸运数
        tagNum = int(str(base)[len(str(base)) -x -1:len(str(base)) -x]) # base比x高一位的数,int型
        tagNum += 1
        answer = (str(base)[:len(str(base)) - x - 1]) + str(tagNum) + str(luckyNum)
        print answer
elif base > luckyNum and len(str(base)) == len(str(luckyNum)):
    # 在luckNum的左面写个1即可了
    print '1' + str(luckyNum)