Java:回文,取不到最大数
Java: Palindrome, not getting biggest number
我想求出3位数中最大的回文数。这是我的代码:
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
int biggest = 1;
String strTemp = temp + "";
if (strTemp.equals(new StringBuilder(strTemp).reverse().toString())) {
if (temp > biggest) {
biggest = temp;
System.out.println("Original: " + strTemp);
System.out.println("Reverse: " + new StringBuilder(strTemp).reverse().toString());
System.out.println("Siffra: " + start);
System.out.println("Siffra2: " + start2);
}
}
}
最后我得到的是995 x 583
而不是993 x 913
,这是最大的。为什么?我有它所以最大的整数总是选择最大的数字。
将 biggest
的声明移到循环外:
int biggest = 1;
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
String strTemp = temp + "";
if (strTemp.equals(new StringBuilder(strTemp).reverse().toString())) {
if (temp > biggest) {
biggest = temp;
System.out.println("Original: " + strTemp);
System.out.println("Reverse: " + new StringBuilder(strTemp).reverse().toString());
System.out.println("Siffra: " + start);
System.out.println("Siffra2: " + start2);
}
}
}
输出:
....
Original: 906609
Reverse: 906609
Siffra: 913
Siffra2: 993
您需要将 int biggest = 1;
移出两个 for 循环。
如果您不在每个内部循环中都这样做,您将重新启动最大的值。
int biggest = 1;
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
String strTemp = temp + "";
if (strTemp.equals(new StringBuilder(strTemp).reverse().toString())) {
if (temp > biggest) {
biggest = temp;
System.out.println("Original: " + strTemp);
System.out.println("Reverse: " + new StringBuilder(strTemp).reverse().toString());
System.out.println("Siffra: " + start);
System.out.println("Siffra2: " + start2);
}
}
}
使用 java 8 你可以重写这段代码如下:
// Define what it means palindrome
IntPredicate isPalindrome = n -> new StringBuffer(String.valueOf(n)).reverse().toString().equals(String.valueOf(n));
OptionalInt max =
// Define a stream from 100 to 1000
IntStream.range(100, 1000)
// Map the original stream to a new stream
// Basically for each x of the first stream
// creates a new stream 100-1000 and map each element
// x of the first stream and y of the second stream
// to x * y
.flatMap(x -> IntStream.range(100, 1000).map(y -> x * y))
// Take only palyndrome of x * y
.filter(isPalindrome)
// take the max
.max();
函数式方法在您必须遍历 n 个元素的大多数情况下更具可读性,并且更容易过滤和提取元素而无需
做错了。
int biggest = 1;
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
在上面你总是initializing biggest=1
。所以你没有得到正确的结果。将它放在 for 循环之外。
我想求出3位数中最大的回文数。这是我的代码:
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
int biggest = 1;
String strTemp = temp + "";
if (strTemp.equals(new StringBuilder(strTemp).reverse().toString())) {
if (temp > biggest) {
biggest = temp;
System.out.println("Original: " + strTemp);
System.out.println("Reverse: " + new StringBuilder(strTemp).reverse().toString());
System.out.println("Siffra: " + start);
System.out.println("Siffra2: " + start2);
}
}
}
最后我得到的是995 x 583
而不是993 x 913
,这是最大的。为什么?我有它所以最大的整数总是选择最大的数字。
将 biggest
的声明移到循环外:
int biggest = 1;
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
String strTemp = temp + "";
if (strTemp.equals(new StringBuilder(strTemp).reverse().toString())) {
if (temp > biggest) {
biggest = temp;
System.out.println("Original: " + strTemp);
System.out.println("Reverse: " + new StringBuilder(strTemp).reverse().toString());
System.out.println("Siffra: " + start);
System.out.println("Siffra2: " + start2);
}
}
}
输出:
....
Original: 906609
Reverse: 906609
Siffra: 913
Siffra2: 993
您需要将 int biggest = 1;
移出两个 for 循环。
如果您不在每个内部循环中都这样做,您将重新启动最大的值。
int biggest = 1;
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
String strTemp = temp + "";
if (strTemp.equals(new StringBuilder(strTemp).reverse().toString())) {
if (temp > biggest) {
biggest = temp;
System.out.println("Original: " + strTemp);
System.out.println("Reverse: " + new StringBuilder(strTemp).reverse().toString());
System.out.println("Siffra: " + start);
System.out.println("Siffra2: " + start2);
}
}
}
使用 java 8 你可以重写这段代码如下:
// Define what it means palindrome
IntPredicate isPalindrome = n -> new StringBuffer(String.valueOf(n)).reverse().toString().equals(String.valueOf(n));
OptionalInt max =
// Define a stream from 100 to 1000
IntStream.range(100, 1000)
// Map the original stream to a new stream
// Basically for each x of the first stream
// creates a new stream 100-1000 and map each element
// x of the first stream and y of the second stream
// to x * y
.flatMap(x -> IntStream.range(100, 1000).map(y -> x * y))
// Take only palyndrome of x * y
.filter(isPalindrome)
// take the max
.max();
函数式方法在您必须遍历 n 个元素的大多数情况下更具可读性,并且更容易过滤和提取元素而无需 做错了。
int biggest = 1;
for (int start = 100; start < 1000; start++) {
for (int start2 = 100; start2 < 1000; start2++) {
int temp = start * start2;
在上面你总是initializing biggest=1
。所以你没有得到正确的结果。将它放在 for 循环之外。