在Java中 - 有什么更快的方法来查找给定的整数是否以数字2开头而不必将数字转换为字符串？
String.valueOf(number).charAt(0) == '2'
答案 0 :(得分：18)
如果您想避免将其转换为字符串，您可以继续除以10以找到最重要的数字：
int getMostSignificantDigit(int x)
{
// Need to handle Integer.MIN_VALUE "specially" as the absolute value can't
// represented. We can hard-code the fact that it starts with 2 :)
x = x == Integer.MIN_VALUE ? 2 : Math.abs(x);
while (x >= 10)
{
x = x / 10;
}
return x;
}
我不知道这是否会比Husman的log / pow方法更快。
答案 1 :(得分：9)
我找到了各种各样的解决方案：
public class FirstDigit
{
static int digit;
@GenerateMicroBenchmark public void string() {
for (int i = 200_000_000; i < 400_000_000; i += 999961)
digit = Integer.toString(i).charAt(0);
}
@GenerateMicroBenchmark public void math() {
for (int i = 200_000_000; i < 400_000_000; i += 999961) {
digit = (int) floor(i / pow(10, floor(log10(i))));
}
}
@GenerateMicroBenchmark public void divide() {
for (int i = 200_000_000; i < 400_000_000; i += 999961) {
int x = i;
while (x > 10) x /= 10;
digit = x;
}
}
@GenerateMicroBenchmark public void brokenDivide() {
for (int i = 200_000_000; i < 400_000_000; i += 999961) {
int x = i;
while (x > 10) x >>= 3;
digit = x;
}
}
@GenerateMicroBenchmark public void bitTwiddling() {
for (int i = 200_000_000; i < 400_000_000; i += 999961) {
digit = i/powersOf10[log10(i)];
}
}
@GenerateMicroBenchmark public boolean avoidDivide() {
boolean b = true;
for (int i = 200_000_000; i < 400_000_000; i += 999961) {
b ^= firstDigitIsTwo(i);
}
return b;
}
private static final int[] log256 = new int[256];
static {
for (int i = 0; i < 256; i++) log256[i] = 1 + log256[i / 2];
log256[0] = -1;
}
private static int powersOf10[] = {1, 10, 100, 1000, 10_000, 100_000,
1_000_000, 10_000_000, 100_000_000, 1_000_000_000};
public static int log2(int v) {
int t, tt;
return ((tt = v >> 16) != 0)?
(t = tt >> 8) != 0 ? 24 + log256[t] : 16 + log256[tt]
: (t = v >> 8) != 0 ? 8 + log256[t] : log256[v];
}
public static int log10(int v) {
int t = (log2(v) + 1) * 1233 >> 12;
return t - (v < powersOf10[t] ? 1 : 0);
}
static final int [] limits = new int[] {
2_000_000_000, Integer.MAX_VALUE,
200_000_000, 300_000_000-1,
20_000_000, 30_000_000-1,
2_000_000, 3_000_000-1,
200_000, 300_000-1,
20_000, 30_000-1,
2000, 3000-1,
200, 300-1,
20, 30-1,
2, 3-1,
};
public static boolean firstDigitIsTwo(int v) {
for ( int i = 0; i < limits.length; i+= 2) {
if ( v > limits[i+1] ) return false;
if ( v >= limits[i] ) return true;
}
return false;
}
}
结果：
Benchmark Mode Thr Cnt Sec Mean Mean error Units
FirstDigit.avoidDivide thrpt 1 3 5 2324.271 58.145 ops/msec
FirstDigit.bitTwiddling thrpt 1 3 5 716.453 6.407 ops/msec
FirstDigit.brokenDivide thrpt 1 3 5 578.259 7.534 ops/msec
o.s.FirstDigit.divide thrpt 1 3 5 125.509 2.323 ops/msec
o.s.FirstDigit.string thrpt 1 3 5 78.233 2.030 ops/msec
o.s.FirstDigit.math thrpt 1 3 5 14.226 0.034 ops/msec
math方法是一个明显的失败者; string方法将math打成6倍; divide算法比这快60％; bitTwiddling算法比divide快六倍; avoidDivide方法（直接给出是/否答案，与所有其他方法不同，实际上确定了第一个数字）是另一个快于bitTwiddiling算法的三倍，是无可争议的赢家; brokenDivide算法。它没有除以十，只是换了三个，给出了错误的答案。关键是要强调divide算法的瓶颈在哪里：brokenDivide比divide慢4.6倍，比bitTwiddling慢0.2倍。请注意，我使用了相当多的数字;相对速度随着幅度而变化。
答案 2 :(得分：8)
我很想做这样的事情：
x = Math.abs(x);
if ( ((int) Math.floor(x / Math.pow(10, Math.floor(Math.log10(x))))) == 2 )
{
... // x equals 2
}
答案 3 :(得分：7)
派生自Bit Twiddling Hacks By Sean Eron Anderson
/*
* Log(2) of an int.
*
* See: http://graphics.stanford.edu/~seander/bithacks.html#IntegerLogLookup
*/
private static final int[] Log256 = new int[256];
static {
for (int i = 0; i < 256; i++) {
Log256[i] = 1 + Log256[i / 2];
}
Log256[0] = -1;
}
public static int log2(int v) {
int t, tt;
if ((tt = v >> 16) != 0) {
return (t = tt >> 8) != 0 ? 24 + Log256[t] : 16 + Log256[tt];
} else {
return = (t = v >> 8) != 0 ? 8 + Log256[t] : Log256[v];
}
}
/*
* Log(10) of an int.
*
* See: http://graphics.stanford.edu/~seander/bithacks.html#IntegerLog10
*/
private static int PowersOf10[] = {1, 10, 100, 1000, 10000, 100000,
1000000, 10000000, 100000000, 1000000000};
public static int log10(int v) {
int t = (log2(v) + 1) * 1233 >> 12;
return t - (v < PowersOf10[t] ? 1 : 0);
}
// Returns the top digit of the integer.
public static int topDigit(int n) {
return n / PowersOf10[log10(n)];
}
我已经对此进行了测试，似乎有效 - 它是如何进行基准测试的？
答案 4 :(得分：4)
更新：我现在在每次迭代时使用不同的数字进行测试，并添加了OldCurmudgeon的解决方案（来自Sean Eron Anderson的Bit Twiddling Hacks的解决方案）。
我对不同的解决方案进行了基准测试，包括Jon Skeet's :)：
以下是我用于此测试的主要方法：
public static void main(String[] args){
long tip = System.currentTimeMillis();
//Call the test method 10 000 000 times.
for(int i= 0 ; i< 10_000_000 ; i++){
//Here I call the method representing the algorithm.
foo3(i);
}
long top = System.currentTimeMillis();
System.out.println("Total time : "+(top-tip));
}
1）使用OP的解决方案：
public static boolean foo1(Integer i){
return String.valueOf(i).charAt(0) == '2';
}
我的电脑上需要 425 ms 。
2）尝试使用Integer.toString（）。startswith（）：
public static boolean foo2(Integer i){
return Integer.toString(i).startsWith("2");
}
我的电脑上需要 410 ms 。
3）借助Husman的解决方案：
public static boolean foo3(Integer i){
i = Math.abs(i);
return ((int) Math.floor(i / Math.pow(10, Math.floor(Math.log10(i))))) == 2;
}
2020 ms 。
4）使用Jon的解决方案：
public static boolean foo4(Integer i){
return getMostSignificantDigit(i)==2;
}
public static int getMostSignificantDigit(int x)
{
// TODO: Negative numbers :)
while (x > 10)
{
x = x / 10;
}
return x;
}
125 ms
5）使用OldCurmudgeon的解决方案（见她的回答）：
public static boolean foo5(Integer i){
return OldCurmudgeon.topDigit(i)==2;
}
97 ms
答案 5 :(得分：4)
还有一种可能性 - 因为分裂显然是一个瓶颈（或者是它？）：
// Pairs of range limits.
// Reverse order to put the widest range at the top.
static final int [] limits = new int[] {
// Can hard-code this one to avoid one comparison.
//2000000000, Integer.MAX_VALUE,
200000000, 300000000-1,
20000000, 30000000-1,
2000000, 3000000-1,
200000, 300000-1,
20000, 30000-1,
2000, 3000-1,
200, 300-1,
20, 30-1,
2, 3-1,
};
public static boolean firstDigitIsTwo(int v) {
// All ints from there up start with 2.
if ( v >= 2000000000 ) return true;
for ( int i = 0; i < limits.length; i += 2 ) {
// Assumes array is decreasing order.
if ( v > limits[i+1] ) return false;
// In range?
if ( v >= limits[i] ) return true;
}
return false;
}
答案 6 :(得分：3)
我在这里回答，所以我可以发布测试代码和结果：
public class NumberStart extends AbstractBenchmark {
private static final int START = 10000000;
private static final int END = 50000000;
private static int result;
@Test
@BenchmarkOptions(benchmarkRounds = 1, warmupRounds = 1)
public void divide() {
for (int x = START; x < END; x++) {
int i = x;
while (i > 10) {
i = i / 10;
}
result = i;
}
}
@Test
@BenchmarkOptions(benchmarkRounds = 1, warmupRounds = 1)
public void math() {
for (int x = START; x < END; x++) {
result = (int) Math.floor(x / Math.pow(10, Math.floor(Math.log10(x))));
}
}
@Test
@BenchmarkOptions(benchmarkRounds = 1, warmupRounds = 1)
public void string() {
for (int x = START; x < END; x++) {
result = (int) Integer.toString(x).charAt(0);
}
}
@Test
@BenchmarkOptions(benchmarkRounds = 1, warmupRounds = 1)
public void bitmath() {
for (int x = START; x < END; x++) {
result = (int) BitMath.topDigit(x);
}
}
}
bitmath()方法使用OldCurmudgeon发布的代码。
以下是结果：
NumberStart.divide: [measured 1 out of 2 rounds, threads: 1 (sequential)]
round: 0.36 [+- 0.00], round.block: 0.00 [+- 0.00], round.gc: 0.00 [+- 0.00], GC.calls: 0, GC.time: 0.00, time.total: 0.71, time.warmup: 0.36, time.bench: 0.35
NumberStart.string: [measured 1 out of 2 rounds, threads: 1 (sequential)]
round: 1.64 [+- 0.00], round.block: 0.00 [+- 0.00], round.gc: 0.00 [+- 0.00], GC.calls: 147, GC.time: 0.08, time.total: 3.31, time.warmup: 1.68, time.bench: 1.64
NumberStart.bitmath: [measured 1 out of 2 rounds, threads: 1 (sequential)]
round: 0.22 [+- 0.00], round.block: 0.00 [+- 0.00], round.gc: 0.00 [+- 0.00], GC.calls: 0, GC.time: 0.00, time.total: 0.45, time.warmup: 0.23, time.bench: 0.22
NumberStart.math: [measured 1 out of 2 rounds, threads: 1 (sequential)]
round: 4.93 [+- 0.00], round.block: 0.00 [+- 0.00], round.gc: 0.00 [+- 0.00], GC.calls: 0, GC.time: 0.00, time.total: 9.89, time.warmup: 4.95, time.bench: 4.93
bitmath()方法是迄今为止最快的方法。
答案 7 :(得分：0)
int intValue=213;
if(String.valueOf(intValue).startsWith("2"))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
答案 8 :(得分：0)
((int) number / 10^(numberlength-1))会给出第一位数字。