我使用HashMap
来存储斐波那契值。
以下是此代码执行的输出:
Enter n: 500000
F(500000) = 2955561408 ... computed in 5,141 ms
Enter n: 500000
F(500000) = 2955561408 ... computed in 0 ms
它缓存很好,并且返回结果很好
我想用更好的Guava cache
替换它,我已经失去了任何利润。
代码执行输出:
Enter n: 500000
F(500000) = 2955561408 ... computed in 5,184 ms
Enter n: 500000
F(500000) = 2955561408 ... computed in 5,086 ms
以下是程序代码:
public class CachedFibonacci {
private static Map<BigDecimal, BigDecimal> previousValuesHolder;
static {
previousValuesHolder = new HashMap<>();
previousValuesHolder.put(BigDecimal.ZERO, BigDecimal.ZERO);
previousValuesHolder.put(BigDecimal.ONE, BigDecimal.ONE);
}
private static LoadingCache<BigDecimal, BigDecimal> cachedFibonacci = CacheBuilder.newBuilder()
.expireAfterWrite(3, TimeUnit.MINUTES)
.maximumSize(500000)
.concurrencyLevel(5)
.weakKeys()
.build(new CacheLoader<BigDecimal, BigDecimal>() {
@Override
public BigDecimal load(BigDecimal key) throws Exception {
return getFibonacciByKey(key);
}
});
private static BigDecimal getFibonacciByKey(BigDecimal key) {
long number = key.longValue();
BigDecimal olderValue = BigDecimal.ONE,
oldValue = BigDecimal.ONE,
newValue = BigDecimal.ONE;
for (int i = 3; i <= number; i++) {
newValue = oldValue.add(olderValue);
olderValue = oldValue;
oldValue = newValue;
}
return newValue;
}
public static BigDecimal getGuavaCache(long number) {
if (0 == number) {
return BigDecimal.ZERO;
} else if (1 == number) {
return BigDecimal.ONE;
} else {
return cachedFibonacci.getUnchecked(BigDecimal.valueOf(number));
}
}
public static BigDecimal getCachedFibonacciOf(long number) {
if (0 == number) {
return BigDecimal.ZERO;
} else if (1 == number) {
return BigDecimal.ONE;
} else {
if (previousValuesHolder.containsKey(BigDecimal.valueOf(number))) {
return previousValuesHolder.get(BigDecimal.valueOf(number));
} else {
BigDecimal olderValue = BigDecimal.ONE,
oldValue = BigDecimal.ONE,
newValue = BigDecimal.ONE;
for (int i = 3; i <= number; i++) {
newValue = oldValue.add(olderValue);
olderValue = oldValue;
oldValue = newValue;
}
previousValuesHolder.put(BigDecimal.valueOf(number), newValue);
return newValue;
}
}
}
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
while (true) {
System.out.print("Enter n: ");
long inputNumber = scanner.nextLong();
if (inputNumber >= 0) {
long beginTime = System.nanoTime();
// BigDecimal fibo = getCachedFibonacciOf(inputNumber);
BigDecimal fibo = getGuavaCache(inputNumber);
long endTime = System.nanoTime();
long delta = endTime - beginTime;
System.out.printf("F(%d) = %.10s ... computed in %,d ms\n", inputNumber, fibo, delta / 1_000_000);
} else {
System.err.println("You must enter number > 0");
System.out.println("try, enter number again, please:");
break;
}
}
}
}
正如我想当你调用cachedFibonacci.getUnchecked()
时,如果它被缓存,它应该返回缓存的值,否则计算它并缓存。
为什么此代码会使用Guava缓存再次计算?
如何解决这个问题?
答案 0 :(得分:5)
如果删除行
.weakKeys()
来自构建缓存,而不是你会看到加速计算。
来自javaDoc
警告:使用此方法时,生成的缓存将使用标识({@code ==})
- 比较以确定键的相等性。