数据类面试题

1，由A-Z、a-z、0-9组成3位的字符密码，设计一个算法，列出并打印所有可能的密码组合。

由于控制台容纳能力有限，故将其写到文本文件中。

没人会耐心等你把这61个字母输完(程序执行ms，输入程序一分钟，自己手动输入四五分钟)，这时用程序把内容生成，然后复制，，，

/*为什么输入程序这么多比手动输入'A','B','C',…'a','b','c'...'1','2,'3'...'8','9'快
*因为需要输入不同的字母，数字，引号括住，还得逗号隔开。
*但是程序都是生成的，输入程序只要半分钟，最多一分钟，执行时间几乎能忽略
*/  
public class ZiFuChuanShengCheng{//这代码程序生成，为方便，可简化为public class A
        public static void main(String[] args){//自动生成
            for(int a='A';a<='Z';a++){//for循环自动生成轮廓，填充即可
		System.out.print("'"+(char)a+"',");
	    }
/*下面两个循环由第一个按住Ctrl+ALT+↓下箭头得到*/
	    for(int a='a';a<='z';a++){
		System.out.print("'"+(char)a+"',");
	    }
	    for(int a='1';a<='9';a++){
		System.out.print("'"+(char)a+"',");
	    }
	    System.out.println();    
    }
 }

　　

　　

　　得到：'A','B','C','D','E','F','G','H','I','J','K','L','M',

　　　　　　'N','O','P','Q','R','S','T','U','V','W','X','Y','Z',

　　　　　　　　'a','b','c','d','e','f','g','h','i','j','k','l','m',

　　　　　　　　　　'n','o','p','q','r','s','t','u','v','w','x','y',

　　　　　　　　　　　　'z','1','2','3','4','5','6','7','8','9',

 

java实现：

 

package com.starain.mimabianli;

import java.io.BufferedWriter;
import java.io.File;
import java.io.FileWriter;
import java.io.IOException;

public class BianLiMiMa {
	public static void main(String[] args) {
	
		try {
			long time1 = System.currentTimeMillis();
			int count = 1;
			char mima[] = {'A','B','C','D','E','F','G','H','I','J','K','L','M',
									'N','O','P','Q','R','S','T','U','V','W','X','Y','Z',
									'a','b','c','d','e','f','g','h','i','j','k','l','m',
									'n','o','p','q','r','s','t','u','v','w','x','y','z',
									'1','2','3','4','5','6','7','8','9'};
			File file1 = new File("D:/mima.txt");
			BufferedWriter bfile1 = new BufferedWriter(new FileWriter(file1));
			
			for(int i = 0;i <61;i++){
				for(int j=0;j<61;j++){
					for(int k = 0; k <61;k++){
						count++;
						bfile1.write(mima[i]);
						bfile1.write(mima[j]);
						bfile1.write(mima[k]);
						bfile1.newLine();
					}
				}
			}
			long time2 = System.currentTimeMillis();
			System.out.println("用时："+(time2-time1)+"ms");
			bfile1.close();
		} catch (IOException e) {
			System.out.println("error");
		}
	
		
		
		
		
	}

}

 

C实现

 1 #include
 2 #include 
 3 void main(){
 4 clock_t t1 , t2;
 5 t1 = clock();
 6 FILE *file1 =fopen("D:/mima2.txt","w");
 7  char mima[] = {
     'A','B','C','D','E','F','G','H','I','J','K','L','M',
 8                 'N','O','P','Q','R','S','T','U','V','W','X','Y','Z',
 9                 'a','b','c','d','e','f','g','h','i','j','k','l','m',
10                 'n','o','p','q','r','s','t','u','v','w','x','y','z',
11                 '1','2','3','4','5','6','7','8','9'};
12  for(int i = 0;i <61;i++){
13     for(int j=0;j<61;j++){
14         for(int k = 0; k <61;k++){
15             fputc(mima[i],file1);
16             fputc(mima[j],file1);
17             fputc(mima[k],file1);
18             fputs("\n",file1);
19             }
20         }
21     }
22  t2 = clock();
23  printf("用时%dms\n",(t2-t1));
24 }

用时真的能可以忽略了。。。

 

2、实现字符串反转函数

String s = "abcdefghijklm";

　C++　

#include 
#include 
using namespace std;

void main(){

    String s = "abcdefghijklm";
    cout <<  s << endl;
    int len = s.length();
    char temp = 'a';
    for(int i = 0; i < len/2; i++){
        temp = s[i];
        s[i] = s[len - 1 - i];
        s[len - 1 - i] = temp;
    }
    
    cout << s;
    
}

　　java

package com.starain.mianshi;

public class FanZhuan {
public static void main(String[] args) {
	String str1 = "abcdefghijklm";
	String str2 = "";
	for(int i=str1.length()-1;i>=0;i--){
		str2+=str1.charAt(i);
	}
	System.out.println(str2);
}
}

　　

转载于:https://www.cnblogs.com/starainDou/p/4709955.html