1.首先将word文档解压缩为zip
/**
* 修改后缀名
*/
public static String reName(String path){
File file=new File(path);
String filename=file.getAbsolutePath();
if(filename.indexOf(".")>=0){
filename=filename.substring(0,filename.lastIndexOf("."));
}
file.renameTo(new File(filename+".zip"));
return filename;
} /**
* 解压缩
*/
public static File zipDeCompressing(String path){
long startTime=System.currentTimeMillis();
File Font=null;
try{
ZipInputStream Zin=new ZipInputStream(new FileInputStream(reName(path)+".zip"));
BufferedInputStream Bin=new BufferedInputStream(Zin); String Parent=reName(path);
ZipEntry entry;
try{
while((entry=Zin.getNextEntry())!=null&&!entry.isDirectory()){
Font=new File(Parent,entry.getName());
if(!Font.exists()){
(new File(Font.getParent())).mkdirs();
}
FileOutputStream out=new FileOutputStream(Font);
BufferedOutputStream Bout=new BufferedOutputStream(out);
int b;
while((b=Bin.read())!=-1){
Bout.write(b);
}
Bout.close();
out.close();
System.out.println(Font+"解压成功");
}
Bin.close();
Zin.close();
}catch(Exception e){
e.printStackTrace();
}
}catch(FileNotFoundException e){
e.printStackTrace();
}
long endTime=System.currentTimeMillis();
System.out.println("耗费时间："+(endTime-startTime)+"ms");
File file=new File(reName(path)+"/word/document.xml");
return file;
}
2.利用Levenshtein Distance (编辑距离)计算文本相似度
private static int min(int one, int two, int three) {
int min = one;
if (two < min) {
min = two;
}
if (three < min) {
min = three;
}
return min;
}
public static int ld(String str1, String str2) {
int d[][]; // 矩阵
int n = str1.length();
int m = str2.length();
int i; // 遍历str1的
int j; // 遍历str2的
char ch1; // str1的
char ch2; // str2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str1
ch1 = str1.charAt(i - 1);
// 去匹配str2
for (j = 1; j <= m; j++) {
ch2 = str2.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[n][m];
}
public static double sim(String str1, String str2) {
try {
double ld = (double)ld(str1, str2);
return (1-ld/(double)Math.max(str1.length(), str2.length()));
} catch (Exception e) {
return 0.1;
}
}源码下载地址：http://download.csdn.net/detail/xiangrikuigt/9696149