亲历迅雷二面

昨天参加了深圳迅雷的二面：

又有一上机题：

读一文本文件：Search.txt

给定一个字符串

算出给定字符串在Txt文件中出现的次数、每行的偏移量、行号和该行内容：

【60分钟题，只用记事本和JDK工具】

 

package com.xunlei;import java.io.BufferedReader;import java.io.File;import java.io.FileInputStream;import java.io.InputStreamReader;import java.util.Scanner;public class SearchTxt { public static void main(String[] args) {  // 读取要比对的子串  System.out.print("请输入要查找的字符串：");  Scanner scanner = new Scanner(System.in);  String subStr = scanner.next();  BufferedReader br = null;  try {   File file = new File("D:/search.txt");   if (!file.exists()) {    file.createNewFile();   }   br = new BufferedReader(new InputStreamReader(new FileInputStream(     file)));   // 定义num   int num = 0;   // 定义行   int row = 0;   // 定义原字符串   String superStr = null;   while ((superStr = br.readLine()) != null) {    row++;    num += findNum(superStr, subStr, row);   }   System.out.println("子串" + subStr + "在Txt文本中出现的次数：" + num);  } catch (Exception e) {  } finally {  } } public static int findNum(String superStr, String subStr, int row) {  char[] ch1 = superStr.toCharArray();  char[] ch2 = subStr.toCharArray();  // 记录次数  int num = 0;  for (int i = 0, j = 0; i < ch1.length; i++) {   if (ch2[j] == ch1[i]) {    j++;    if (j == ch2.length) {     num++;     // 第几行有相同的子串     System.out.println("第" + row + "行：" + superStr);     // 偏移量为     System.out.println("偏移量为：" + (i - j + 2));     // i重置为 (i-j+1)     i = i - j + 1;     // j重置为0     j = 0;    }   } else {    j = 0;   }  }  return num; }}