快手算法笔试题目大题2019-3-30

第一题

选择一个数里面出现的奇次数和偶次数的个数

num = int(input())
list1 = [int(tmp) for tmp in input().strip().split(" ")]
jishu = []
oshu = [] for i in range(0,len(list1)): if list1[i] in jishu:
        oshu.append(list1[i])
        jishu.remove(list1[i]) elif list1[i] in oshu:
        jishu.append(list1[i])
        oshu.remove(list1[i]) else:
        jishu.append(list1[i]) for i in jishu: print(i)

第二题

判断是否是递增的序列或者修改一个数让他是递增的序列

list1 = [int(tmp) for tmp in input().strip().split(" ")] def check(list1): for i in range(len(list1)-1): if list1[i] > list1[i+1]: return -1  return 1   def checklist(list1): for i in range(len(list1) - 1): if list1[i] > list1[i + 1]: if (i == 0):
                list1[i] = list1[i + 1] else:
                list1[i] = list1[i - 1]
            res = check(list1) if (res == -1): print(0) else: print(1) break count = 0 for i in range(len(list1)-1): if list1[i] < list1[i+1]:
        count = count+1  if count == len(list1)-1: print(1) else:
    checklist(list1)

第三题

实现一个计算器功能使得 + -* () 满足

python

import re import sys def format_input(input_re_value):
    input_re_value = input_re_value.replace(" ", "") #去掉输入的空格  input_re_value = re.split(r'(\D)', input_re_value) while True: if "" in input_re_value:
          input_re_value.remove("") #将列表中""去掉  else: break   return input_re_value # 计算没有括号的列表的值。加减乘除 def comput(re_value): while "*" in re_value or "/" in re_value: for i, j in enumerate(re_value): if j == "*":
                re_cheng = int(re_value[i - 1]) * int(re_value[i + 1]) #re_cheng = re_value[i - 1] * re_value[i + 1]  re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.insert(i - 1, re_cheng) break   if j == "/":
                re_chu = int(re_value[i - 1]) / int(re_value[i + 1])
                re_chu = int(re_chu)
                re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.insert(i - 1, re_chu) break  while "+" in re_value or "-" in re_value: for i, j in enumerate(re_value): if j == "+":
                re_jia = int(re_value[i - 1]) + int(re_value[i + 1])
                re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.insert(i - 1, re_jia) break   if j == "-":
                re_jian = int(re_value[i - 1]) - int(re_value[i + 1])
                re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.pop(i - 1)
                re_value.insert(i - 1, re_jian) break  return re_value[0] def bracket(input_str): if not re.search(r'\(([^()]+)\)', input_str): #若没有括号  s = format_input(input_str) return comput(s) else: #有括号  li = re.split(r'\(([^()]+)\)', input_str)
        data = re.search(r'\(([^()]+)\)', input_str).group()
        data_strip = data.strip("()")
        inde = li.index(data_strip) #取得其下标   ret = comput(format_input(data_strip)) #参数为列表形式  li.pop(inde)
        li.insert(inde, str(ret))
        re_str = "".join(li) if "+" not in re_str and "-" not in re_str and "*" not in re_str \ and "/" not in re_str and "(" not in re_str and ")" not in re_str: return re_str #为什么没有return  else: return bracket(re_str) #迭代    input = input()
res = bracket(input) print(res)

c语言

#include <stdio.h>
#include <stdlib.h>
int main()
{
int x, y;
char input;
int ret = 0;

scanf("%d%c%d", &x,&input,&y);
switch (input)
{
case'+':
ret = x + y;
break;
case'-':
ret = x - y;
break;
case'*':
ret =x * y;
break;
default:
break;
}
printf("%d\n", ret);

return 0;
}

快手 算法笔试题目大题2019-3-30

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快手算法笔试题目大题2019-3-30