”让男友打十万个「对不起」道歉，汉字标上多少遍。如何快速检查中间没有漏？“的坚决方案

import java.util.Stack;

public class Main {

static String[] values = {"零", "一", "二", "三", "四", "五", "六", "七", "八", "九"};
static String[] pos = {"拾", "佰", "仟", "万"};
private static Stack<String> mStack;

public static void main(String[] args) {
long start = System.currentTimeMillis();
for (int i = 1; i <= 99999; i++) {
if (i >= 10 && i < 20) {
System.out.println("对不起 第" + change(i).substring(1) + "遍");
} else {
System.out.println("对不起 第" + change(i) + "遍");
}
}
long end = System.currentTimeMillis();
System.out.println("运行时间" + (end - start) + "ms");
}

private static String change(int i) {
boolean flag = false;
int len = 0;
int num = i;
while (i > 0) {
i /= 10;
len++;
}
if (len > 1) {
mStack = new Stack<String>();
for (int j = 1; j < len; j++) {
if (numAtInteger(num, j) != 0)
mStack.push(pos[j - 1]);
}
}
StringBuilder sb = new StringBuilder();
for (int j = len - 1; j > 0; j--) {
int temp = (num / ((int) Math.pow(10, j)));
num %= ((int) Math.pow(10, j));
if (temp != 0) {
sb.append(values[temp]);
sb.append(mStack.pop());
}
if (temp == 0 && !flag) {
sb.append("零");
flag = true;
}
}
if (num == 0) {
if (sb.toString().endsWith("零")) {
return sb.substring(0, sb.length() - 1);
}
return sb.toString();
} else {
sb = sb.append(values[num]);
return sb.toString();
}
}
static int numAtInteger(int num, int j) {
while (j > 0) {
num /= 10;
j--;
}
return num % 10;
}
}