笔试算法题及解答（Python）

Posted 2021-01-20 hugh.zm

1、给定任意一个正整数，求比这个数大且最小的“不重复数”，“不重复数”的含义是相邻两位不相同，例如1101是重复数，而1201是不重复数

 1 # coding:utf-8
 2 ‘‘‘
 3 Created on 2015年8月11日
 4 Q1: 给定任意一个正整数，求比这个数大且最小的“不重复数”，“不重复数”的含义是相邻两位不相同，例如1101是重复数，而1201是不重复数
 5 @author: Hongzm
 6 ‘‘‘
 7 
 8 def judgeRepetNum(numArr):
 9     size = len(numArr)
10     for i in range(size-1):
11         if(numArr[i] == numArr[i+1]):
12             return True
13     return False
14 
15 if __name__ == ‘__main__‘:
16     num = 112340
17     while(num<9900000000):
18         numArr = list(str(num))
19         if not judgeRepetNum(numArr):
20             print num
21             break
22         else:
23             num += 1

2、长度为N(N很大)的字符串，求这个字符串里的最长回文子串

# coding:utf-8
‘‘‘
Created on 2015年8月11日
Q2: 长度为N(N很大)的字符串，求这个字符串里的最长回文子串
@author: Hongzm
‘‘‘
import re

# manacher算法
def manacherStr(str):
    # p记录每个节点回文子串半径
    p = []
    mx = 0
    id = 0 
    size = len(str)
    for i in range(1, size-1):
        p[i] = min(p[2*id-i], mx -1) if mx>i else 1
        while(str[i + p[i]] == str[i-p[i]]):
            p[i]+=1
        if(i + p[i]>mx):
            mx = i + p[i]
            id = i
    print p

def palindromeStr(words):
    print "start"
    maxLen = 0
    index = 0
    size = len(words)
    for i in range(1, size-1):
        print "i = ", i, words[i]
        tempLen = 0
        for j in range(1, min(i, size-i)):
            print "j = ", j
            if(words[i-j]==words[i+j]):
                    tempLen += 1
            else:
                break
        print "tempLen = ", tempLen
        if(tempLen > maxLen):
            maxLen = tempLen
            index = i
        print "maxLen = ", maxLen
    return words[index-maxLen:index+maxLen+1]                

if __name__ == ‘__main__‘:
    str_origin = "fabbacddcabh"
    str_ = "-" + "-".join(str_origin) + "-"
    str_end = palindromeStr(str_)
    print str_end.replace(‘-‘,"")
    print re.sub("python", "word", "hello, python")

3、数轴上从左到右有n各点a[0], a[1], ……,a[n -1]，给定一根长度为L的绳子，求绳子最多能覆盖其中的几个点

# coding:utf-8

‘‘‘
Created on 2015年8月11日
Q3、数轴上从左到右有n各点a[0], a[1], ……,a[n -1]，给定一根长度为L的绳子，求绳子最多能覆盖其中的几个点
@author: Hongzm
‘‘‘

theLen = 26
oriArr = [-20, -10, -5, 0 , 1, 6, 10, 18, 25]
lenArr = []
oriSize = len(oriArr)
for i in range(oriSize-1):
    lenArr.append(oriArr[i+1] - oriArr[i])
# print lenArr

maxNum = 0
index = 0
lenSize = len(lenArr)
for i in range(lenSize):
    tempLen = 0
    temp = 0
    j= 0
    while(tempLen <= theLen and j< lenSize-i):
        temp +=1
        tempLen += lenArr[i+j]
        j +=1
    if(temp > maxNum):
        maxNum = temp
        index = i

print maxNum, index
print oriArr[index:index+maxNum]