微软2014实习生及秋令营技术类职位在线测试_题目4 : Most Frequent Logs

时间限制:10000ms
单点时限:3000ms
内存限制:256MB

Description

In a running system, there are many logs produced within a short period of time, we'd like to know the count of the most frequent logs.

Logs are produced by a few non-empty format strings, the number of logs is N(1<=N<=20000), the maximum length of each log is 256.

Here we consider a log same with another when their edit distance (see note) is <= 5.

Also we have a) logs are all the same with each other produced by a certain format string b) format strings have edit distance  5 of each other.

Your program will be dealing with lots of logs, so please try to keep the time cost close to O(nl), where n is the number of logs, and l is the average log length.

Note edit distance is the minimum number of operations (insertdeletereplace a character) required to transform one string into the other, please refer to
http://en.wikipedia.org/wiki/Edit_distance for more details.

Input

Multiple lines of non-empty strings.

Output

The count of the most frequent logs.

样例输入

Logging started for id:1
Module ABC has completed its job
Module XYZ has completed its job
Logging started for id:10
Module ? has completed its job

样例输出

3

//source here

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main{

static int count = 0,maxcount = 0;
static List<String> list = new ArrayList<String>();

public static void main(String[] args) {

int N = 0;
int i=0,j=0,k=0;
Boolean flag = false;
@SuppressWarnings("resource")
Scanner in = new Scanner(System.in);//不会检测输入越界的问题，但是注意整型上限，最好将变量定义为long长整型
do{
String line = in.nextLine();
if( line==null || "".equals(line.toString())){
flag = true;
}else{
list.add(line);
}
}while(!flag && N<20000);

for(i=0;i<list.size()-1;i++){
for(j=i+1;j<list.size();j++){
String strA = list.get(i);
String strB = list.get(j);
int result=edit_distance_Result(strA, strB);
if(result <= 5){
count++;
}
}
if(maxcount<count){
maxcount = count;
}
count = 0;
}
System.out.println(++maxcount);
}

/**

* 相似度比较

* @param strA

* @param strB

* @return

*/

public static int edit_distance_Result(String strA, String strB){

String newStrA = removeSign(strA);

String newStrB = removeSign(strB);

int temp = Math.max(newStrA.length(), newStrB.length());

int temp2 = compare(newStrA, newStrB); //比较两个字符串，返回未匹配的字符串个数

//System.out.println("temp2 == "+temp2);

return (temp2);

}

private static int compare(String str, String target) {//
int d[][]; // 矩阵
int n = str.length();
int m = target.length();
int i; // 遍历str的
int j; // 遍历target的
char ch1; // str的

char ch2; // target的

int temp; // 记录相同字符,在某个矩阵位置值的增量,不是0就是1
if (n == 0) {
return m;
}

if (m == 0) {
return n;
}

d = new int[n + 1][m + 1];
for (i = 0; i <= n; i++) { // 初始化第一列

d[i][0] = i;

}
for (j = 0; j <= m; j++) { // 初始化第一行

d[0][j] = j;

}
for (i = 1; i <= n; i++) {// 遍历str
ch1 = str.charAt(i - 1);
// 去匹配target
for (j = 1; j <= m; j++){

ch2 = target.charAt(j - 1);
if (ch1 == ch2) {
temp = 0;
} else {

temp = 1;
}
// 左边+1,上边+1, 左上角+temp取最小
d[i][j] = min(d[i - 1][j] + 1, d[i][j - 1] + 1, d[i - 1][j - 1] + temp);
}
}

return d
[m];

}

private static int min(int one, int two, int three) {
return (one = one < two ? one : two) < three ? one : three;

}

private static String removeSign(String str) {

StringBuffer sb = new StringBuffer();

for (char item : str.toCharArray())

if (charReg(item)){

//System.out.println("--"+item);

sb.append(item);

}

return sb.toString();

}

private static boolean charReg(char charValue) {

return (charValue >= 0x4E00 && charValue <= 0X9FA5)

|| (charValue >= 'a' && charValue <= 'z')

|| (charValue >= 'A' && charValue <= 'Z')

|| (charValue >= '0' && charValue <= '9');

}

/*

private static String longestCommonSubstring(String strA, String strB) {

char[] chars_strA = strA.toCharArray();

char[] chars_strB = strB.toCharArray();

int m = chars_strA.length;

int n = chars_strB.length;

int[][] matrix = new int[m + 1][n + 1];

for (int i = 1; i <= m; i++) {

for (int j = 1; j <= n; j++) {

if (chars_strA[i - 1] == chars_strB[j - 1])

matrix[i][j] = matrix[i - 1][j - 1] + 1;

else

matrix[i][j] = Math.max(matrix[i][j - 1], matrix[i - 1][j]);

}

}

char[] result = new char[m+n+1];

int currentIndex = result.length - 1;

while (matrix
.length != 0 && m>1 && n>1) {

if (matrix
== matrix[n - 1])

n--;

else if (matrix
.length == matrix[m - 1]
)

m--;

else {

result[currentIndex] = chars_strA[m - 1];

currentIndex--;

n--;

m--;

}

}

return new String(result);

}*/

}

这是作者提供的一个答案，但是未经过ms hihocoder.平台的测试，因为时间太短，来不及提交！可以输出结果，时间空间的效率还不高，欢迎大家一起讨论！