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锁定老帖子 主题：一道简单的面试题
该帖已经被评为新手帖
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发表时间：2010-04-05
乱七八糟的呀。。。这样的优化只能说是“硬”优化吧。。
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发表时间：2010-04-05
学习了，感觉优化还是有效果的饿，但是如果不是太大的循环的话必要性不是太大
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发表时间：2010-04-05
laoshifu 写道 实例化次数计算不对，对于java来说，for语句的初始化只是执行一次。所以对于原代码，优化一和优化二实例化变量的实例化次数都是一。不对之处，多多指教。
for语句的初始化只是执行一次，循环体执行不只一次，所以第2、3层循环不只初始化一次。 |
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发表时间：2010-04-06
for (int i = 0; i < 1000; i++)
for (int j = 0; j < 100; j++) for (int k = 0; k < 10; k++) log(i * j * k); i*j*k 当i=0 或者j=0, 或者k=0时，，都相当于log(0),,也就是log(0)重复计算了1000×100×10次，，，类似的重复计算太多 优化点应该是把log的重复计算降低 |
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发表时间：2010-04-06
zhangkaitao 写道 测试机器 CPU P8700 @ 2.53GHZ MEMORY 2GB
public static void testFor0() {
int count = 0;
int i = 1;
int j = 1;
int k = 1;
long startTime = System.nanoTime();
for (; i <= 100; i++)
for (j = 1; j <= 1000; j++)
for (k = 1; k <= 10000; k++) {
count++;
}
System.out.println("zero count==" + count);
long endTime = System.nanoTime();
System.out.println("zero time==== " + (endTime - startTime));
}
public static void testFor1() {
int count = 0;
int i = 1;
int j = 1;
int k = 1;
long startTime = System.nanoTime();
for (; i <= 10000; i++)
for (j = 1; j <= 1000; j++)
for (k = 1; k <= 100; k++) {
count++;
}
System.out.println("first count==" + count);
long endTime = System.nanoTime();
System.out.println("first time==== " + (endTime - startTime));
}
public static void testFor2() {
int count = 0;
int i = 1;
int j = 1;
int k = 1;
long startTime = System.nanoTime();
for (; i <= 100; i++)
for (j = i; j <= 1000; j++)
for (k = j; k <= 10000; k++) {
count++;
}
System.out.println("second count==" + count);
long endTime = System.nanoTime();
System.out.println("second time==== " + (endTime - startTime));
}
public static void testFor3() {
int ii = 10000, jj = 1000, kk = 100, sum = ii * jj * kk, iii = 0, kkk = 0, jjj = 0;
int count = 0;
long startTime = System.nanoTime();
for (ii = 0, jj = 0, kk = 0; kk < sum; kk++) {
kkk = kk % 100;
if (kkk == 0 && kk > 0) {
jj++;
jjj = jj % 1000;
if (jjj == 0) {
ii++;
iii = ii%10000;
}
}
count++;
}
System.out.println("third count==" + count);
long endTime = System.nanoTime();
System.out.println("third time==== " + (endTime - startTime));
}
结果 zero count==1000000000 zero time==== 858950788 first count==1000000000 first time==== 941317961 second count==900711700 second time==== 756062573 third count==1000000000 third time==== 6520663101
public static void testFor0() {
int count = 0;
int i = 1;
int j = 1;
int k = 1;
long startTime = System.nanoTime();
for (; i <= 10; i++)
for (j = 1; j <= 100; j++)
for (k = 1; k <= 1000; k++) {
count++;
}
System.out.println("zero count==" + count);
long endTime = System.nanoTime();
System.out.println("zero time==== " + (endTime - startTime));
}
public static void testFor1() {
int count = 0;
int i = 1;
int j = 1;
int k = 1;
long startTime = System.nanoTime();
for (; i <= 1000; i++)
for (j = 1; j <= 100; j++)
for (k = 1; k <= 10; k++) {
count++;
}
System.out.println("first count==" + count);
long endTime = System.nanoTime();
System.out.println("first time==== " + (endTime - startTime));
}
public static void testFor2() {
int count = 0;
int i = 1;
int j = 1;
int k = 1;
long startTime = System.nanoTime();
for (; i <= 10; i++)
for (j = i; j <= 100; j++)
for (k = j; k <= 1000; k++) {
count++;
}
System.out.println("second count==" + count);
long endTime = System.nanoTime();
System.out.println("second time==== " + (endTime - startTime));
}
public static void testFor3() {
int ii = 1000, jj = 100, kk = 10, sum = ii * jj * kk, iii = 0, kkk = 0, jjj = 0;
int count = 0;
long startTime = System.nanoTime();
for (ii = 0, jj = 0, kk = 0; kk < sum; kk++) {
kkk = kk % 10;
if (kkk == 0 && kk > 0) {
jj++;
jjj = jj % 100;
if (jjj == 0) {
ii++;
iii = ii%1000;
}
}
count++;
}
System.out.println("third count==" + count);
long endTime = System.nanoTime();
System.out.println("third time==== " + (endTime - startTime));
}
结果 zero count==1000000 zero time==== 1811683 first count==1000000 first time==== 2361474 second count==905620 second time==== 2054730 third count==1000000 third time==== 7406249 大家看看结果吧。 有理有据，看样子10 100 1000的循环优化还是有效果的。 |
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发表时间：2010-04-06
听说递减效率更好
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发表时间：2010-04-06
怎么还不如原来的了呢.......
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发表时间：2010-04-06
1.i,j,k必须从1开始
2.i*j*k是有重复的 3.log(i*j*k)=logi+logj+logk 这才是这道题的关键吧，像这种初始化的东西，根本相差不大 |
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发表时间：2010-04-06
int i, j, k, m;
for (i = 0; i < 10; i++){
for (j = 0; j < 100; j++){
m=i*j;
for (k = 0; k < 100; k++){
log(m * k);
}
}
}
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发表时间：2010-04-06
感觉差不多，现在的服务器一般内存都很大，不会在乎那么一丁点的损耗
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