问题描述
我正在尝试查找1到10000000（包括两端）之间的所有数字。我尝试了两种解决方案
- 暴力法：循环遍历从1到10000000的所有数字，并找到可被3或5或两者整除的所有数字。
- “分而治之”的方法：具有4个计数器（起始处2个，结束处2个）。 2个计数器工作于3的倍数，两个工作于5的倍数。我将所有倍数放在一个集合中（我不需要Sorted元素，我只需要Element，排序也增加了我的复杂性）。
但是，循环方法比“分而治之”方法花费的时间要少（大约少10倍）。 我也在网上搜索解决方案。但是，我只能找到循环方法。我的方法中缺少什么东西，这增加了我的执行时间吗？请指出我这一点。我从一个列表开始，移到“排序集”，然后最终决定使用HashSet，但似乎需要时间。
这是我尝试过的。
`
public static void main(String[] args) {
System.out.println("Numbers divisible by 3 and 5:");
nosDivisibleBy3And5(); // divide & conquer approach (approach to consider)
nosDivisibleBy3And5BruteForce();
}
private static void nosDivisibleBy3And5BruteForce() {
IntStream ar = IntStream.range(1,10000001); // start inclusive,end exclusive
Integer[] array = ar.boxed().toArray(Integer[]::new);
List<Integer> list = new ArrayList<>();
int count = 0;
long start = System.currentTimeMillis();
/*
* Traversing array from 1 to 100,* if it is either divisible by 3 or 5 or both,count it,print it.
*
*/
for(int i = 0; i < array.length ; i ++) {
if((array[i] % 3 == 0) || (array[i] % 5 == 0)) {
//System.out.println(array[i]);
list.add(array[i]);
count++;
}
}
long end = System.currentTimeMillis();
System.out.println("Brute Force Approach:");
System.out.println("No of elements counted: " + count);
//Collections.sort(list);
//System.out.println("Elements: " + list);
System.out.println("Time: " + (end - start));
}
private static void nosDivisibleBy3And5() {
/*
* Set has all those numbers which
* are divisible by both 3 and 5.
*
*/
Set<Integer> elementsSet = new HashSet<Integer>();
int fr3,fr5,mid,count;
fr3 = 2; // fr3 indicates the index of the first value divisible by 3.
fr5 = 4; // fr5 indicates the index of the first value divisible by 5.
count = 0;
int end3 = 9999998,// end3 indicates the index of the last value divisible by 3.
end5 = 9999999; // end5 indicates the index of the last value divisible by 5.
/* Getting all the numbers from 1 to 100 from Intstream object */
IntStream ar = IntStream.range(1,end exclusive
Integer[] array = ar.boxed().toArray(Integer[]::new);
/*
* Using divide and conquer approach,mid divides the array from 1 to 100
* in two parts,on the first fr3 and fr5 will work,on the second part end3
* and end5 will work.
*/
mid = (fr3 + end3)/2;
long start = System.currentTimeMillis();
while(fr3 <= mid && end3 >= mid) {
elementsSet.add(array[fr3]);
elementsSet.add(array[fr5]);
elementsSet.add(array[end3]);
elementsSet.add(array[end5]);
fr3 += 3;
fr5 += 5;
end3 -= 3;
end5 -= 5;
}
long end = System.currentTimeMillis();
System.out.println("Our approach");
System.out.println("No of elements counted: " + elementsSet.size());
//System.out.println("Elements:" + elementsSet);
System.out.println("Time: " + (end - start));
}
}
`
解决方法
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