java XXTEA加密算法的实现

jgszhuzhu 2012-10-18 10:15:54
求java XXTEA加密算法的实现
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kosora曹 2012-10-19
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好吧,我表示没听说过,恕我无能为力,呵呵。
jgszhuzhu 2012-10-18
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你这个是抄袭的,我要的是具体实现,我自己写出来了
kamouswjw 2012-10-18
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/* xxtea.java
*
* Author: Ma Bingyao < andot@ujn.edu.cn >
* Copyright: CoolCode.CN
* Version: 1.0
* LastModified: 2006-05-11
* This library is free. You can redistribute it and/or modify it.
* http://www.coolcode.cn/?p=169
*/
public class xxtea {
/**
* Encrypt data with key.
*
* @param data
* @param key
* @return
*/
public static byte [] encrypt ( byte [] data , byte [] key ) {
if ( data . length == 0 ) {
return data ;
}
return toByteArray ( encrypt ( toIntArray ( data , true ) , toIntArray ( key , false )) , false ) ;
}

/**
* Decrypt data with key.
*
* @param data
* @param key
* @return
*/
public static byte [] decrypt ( byte [] data , byte [] key ) {
if ( data . length == 0 ) {
return data ;
}
return toByteArray ( decrypt ( toIntArray ( data , false ) , toIntArray ( key , false )) , true ) ;
}

/**
* Encrypt data with key.
*
* @param v
* @param k
* @return
*/
public static int [] encrypt ( int [] v , int [] k ) {
int n = v . length - 1 ;
if ( n < 1 ) {
return v ;
}
if ( k . length < 4 ) {
int [] key = new int [ 4 ] ;
System . arraycopy ( k , 0 , key , 0 , k . length ) ;
k = key ;
}
int z = v [ n ] , y = v [ 0 ] , delta = 0x9E3779B9 , sum = 0 , e ;
int p , q = 6 + 52 / ( n + 1 ) ;
while ( q -- > 0 ) {
sum = sum + delta ;
e = sum >>> 2 & 3 ;
for ( p = 0 ; p < n ; p ++ ) {
y = v [ p + 1 ] ;
z = v [ p ] += ( z >>> 5 ^ y << 2 ) + ( y >>> 3 ^ z << 4 ) ^ ( sum ^ y ) + ( k [ p & 3 ^ e ] ^ z ) ;
}
y = v [ 0 ] ;
z = v [ n ] += ( z >>> 5 ^ y << 2 ) + ( y >>> 3 ^ z << 4 ) ^ ( sum ^ y ) + ( k [ p & 3 ^ e ] ^ z ) ;
}
return v ;
}

/**
* Decrypt data with key.
*
* @param v
* @param k
* @return
*/
public static int [] decrypt ( int [] v , int [] k ) {
int n = v . length - 1 ;
if ( n < 1 ) {
return v ;
}
if ( k . length < 4 ) {
int [] key = new int [ 4 ] ;
System . arraycopy ( k , 0 , key , 0 , k . length ) ;
k = key ;
}
int z = v [ n ] , y = v [ 0 ] , delta = 0x9E3779B9 , sum , e ;
int p , q = 6 + 52 / ( n + 1 ) ;
sum = q * delta ;
while ( sum != 0 ) {
e = sum >>> 2 & 3 ;
for ( p = n ; p > 0 ; p -- ) {
z = v [ p - 1 ] ;
y = v [ p ] -= ( z >>> 5 ^ y << 2 ) + ( y >>> 3 ^ z << 4 ) ^ ( sum ^ y ) + ( k [ p & 3 ^ e ] ^ z ) ;
}
z = v [ n ] ;
y = v [ 0 ] -= ( z >>> 5 ^ y << 2 ) + ( y >>> 3 ^ z << 4 ) ^ ( sum ^ y ) + ( k [ p & 3 ^ e ] ^ z ) ;
sum = sum - delta ;
}
return v ;
}

/**
* Convert byte array to int array.
*
* @param data
* @param includeLength
* @return
*/
private static int [] toIntArray ( byte [] data , boolean includeLength ) {
int n = ((( data . length & 3 ) == 0 ) ? ( data . length >>> 2 )
: (( data . length >>> 2 ) + 1 )) ;
int [] result ;
if ( includeLength ) {
result = new int [ n + 1 ] ;
result [ n ] = data . length ;
} else {
result = new int [ n ] ;
}
n = data . length ;
for ( int i = 0 ; i < n ; i ++ ) {
result [ i >>> 2 ] |= ( 0x000000ff & data [ i ]) << (( i & 3 ) << 3 ) ;
}
return result ;
}

/**
* Convert int array to byte array.
*
* @param data
* @param includeLength
* @return
*/
private static byte [] toByteArray ( int [] data , boolean includeLength ) {
int n ;
if ( includeLength ) {
n = data [ data . length - 1 ] ;
} else {
n = data . length << 2 ;
}

byte [] result = new byte [ n ] ;
for ( int i = 0 ; i < n ; i ++ ) {
result [ i ] = ( byte ) ( data [ i >>> 2 ] >>> (( i & 3 ) << 3 )) ;
}
return result ;
}
}

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