//根据不同的难度产生随机字母和数字
for(int i=0; readomNumStart.length()<4; i++) {
if(difficult == 1) {
//产生随机的0-9的数字
a = String.valueOf((int)(Math.random() * 10)) ;
}
if(difficult == 2) {
//在0-9和a，b，c，d中随机产生。
String[] readomWord = {"0","1","2","3","4","5","6","7","8","9","a","b","c","d"};
int readomWordIndex = (int)(Math.random() * 13);
a = readomWord[readomWordIndex];
}
if(difficult == 3) {
//产生随机的字母
char readomLetter = (char)(Math.random ()*26+'a');
a = String.valueOf(readomLetter) ;
}
if(difficult == 4) {
//产生随机的数字和字母的组合
String[] readomHard = new String[20];
int readomWordIndex = (int)(Math.random() * 19);
for(int j=0; j<20; j++) {
int readomWordNum = (int)(Math.random() * 10);
char readomLetter = (char)(Math.random ()*26+'a');
if(readomWordNum % 2 == 0) {
readomHard[j] = readomWordNum + "";
}else{
readomHard[j] = String.valueOf(readomLetter);
}
}
a = readomHard[readomWordIndex];
}
// |
这样写代码比较短。我也参考了许多人的代码大部分都很复杂。我不喜欢写那么多代码，所以就这样写了。还不能知道这样写效率有没有问题，在我本机上运行没有什么感觉。
posted on 2009-04-07 08:09
Werther 阅读(3341)
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10.Java