所以，对于类，我必须编写一个程序来确定有多少正整数
1）100万以下
2）数字中至少有一个7和一个9
3）必须使用蛮力法。
虽然答案应该是199262，但由于重复，我一直得到228530，有人能看看我哪里出错了吗谢谢！
相似但不相同的问题：Java - numbers with at least one 7 and one 9 in its digit
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i<1000000; i++) //Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); //sets a nmber to a string
int length = number2.length() -1; //length goes up to the last character 0-j
for (int j= 0; j <= length; j++) //looking for the first 7 or 9 in string
{
char a = number2.charAt(j); //sets char to the next "letter"
if (a == '7' && oneSeven != true) //if the number is a 7 and there isnt already a seven
{
oneSeven = true; //now there is a seven,
for (int k = j+1; k <= length; k++) //checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9')
{
sevNine = true;
}
}
}
else if (a == '9' && oneNine != true)
{
oneNine = true;
for (int l = j+1; l <= length; l++)
{
char b = number2.charAt(l);
if (b == '7')
{
sevNine = true;
}
}
}
if (sevNine == true)
{
counter++;
sevNine = false;
System.out.println(number2);
}
}
}
System.out.println(counter);
最佳答案
一旦sevNine设置为true并递增计数器，则不会中断循环，因此它会在同一个数字上的每个数字上迭代，即使它已经包含该数字…只需添加一个中断语句即可在每次递增计数器后退出每个循环上的for循环迭代。
这是密码。
public static void main(String[] args) {
boolean sevNine = false; // a combination of seven and nine in a number
boolean oneNine;
boolean oneSeven;
int counter = 0;
for (int i = 0; i < 1000000; i++) // Runs numbers 1-1000000
{
oneSeven = false;
oneNine = false;
String number2 = " " + (i); // sets a nmber to a string
int length = number2.length() - 1; // length goes up to the last character 0-j
for (int j = 0; j <= length; j++) // looking for the first 7 or 9 in string
{
char a = number2.charAt(j); // sets char to the next "letter"
if (a == '7' && oneSeven != true) // if the number is a 7 and there isnt already a seven
{
oneSeven = true; // now there is a seven,
for (int k = j + 1; k <= length; k++) // checks from the next char up to the length for a 9
{
char b = number2.charAt(k);
if (b == '9') {
sevNine = true;
}
}
} else if (a == '9' && oneNine != true) {
oneNine = true;
for (int l = j + 1; l <= length; l++) {
char b = number2.charAt(l);
if (b == '7') {
sevNine = true;
}
}
}
if (sevNine == true) {
counter++;
sevNine = false;
System.out.println(number2);
break;
}
}
}
System.out.println(counter);
}
如果使用break语句运行，则应得到
199262作为结果数。