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;
 }
}