public static void sort(Object[] objects){
if(objects instanceof Number[]){
for (int i = 0; i < objects.length-1; i++) {
for (int j = i+1; j < objects.length; j++) {
//判断前一个值是否大于后一个值
if(((Number)objects[i]).doubleValue()>((Number) objects[j]).doubleValue()){
Object temp = objects[i];
objects[i] = objects[j];
objects[j] = temp;
}
}
}
}else if(objects instanceof String[]){
for (int i = 0; i < objects.length-1; i++) {
for (int x = i+1; x < objects.length; x++) {
//拿到前一个String
char[] charBefore = ((String)objects[i]).toCharArray();
//拿到后一个String
char[] charsAfter = ((String)objects[x]).toCharArray();
//计算当前变换的String[x]的长度
int length = ((String) objects[x]).length();
//获取最短循环次数,以免下标越界
int loopNumber = ( charBefore.length > length ? length : charBefore.length );
for (int j = 0; j < loopNumber; j++) {
//判断前一位的首字母是否小于后一位,例如:首字母a<首字母b则不用再比较
if((int)charBefore[j]<(int)charsAfter[j]){
break;
}
//判断前一位字母是否大于后一位字母,如果大于则替换位置
else if((int)charBefore[j]>(int)charsAfter[j]){
Object temp = objects[i];
objects[i] = objects[x];
objects[x] = temp;
break;
}
//如果两个String的长度不同,最短内容的最短起始位到最短结束位 和 最长内容的最短起始位和最短结束位的这段字符串长度内容相等,
// 则长度最短的放在前面,例如String a = "asd";String b = "as"; 则 字符串b排在前面
else if (j == loopNumber-1 && charBefore[loopNumber-1] == charsAfter[loopNumber-1]){
if(charBefore.length > charsAfter.length) {
Object temp = objects[i];
objects[i] = objects[x];
objects[x] = temp;
}
}
}
}
}
}
}