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1524450592
tags: Hash Table, Math
计数: 所有小于n的prime number.
#### prime number定义
- >=2的没有除自己和1以外公约数的数。
- 还有另外一个定义方法: 这个n,有没有小于n的一个i, 而达到： i * i + # of i = n. 如果有，那就不是 prime
#### Steps
- 一个boolean长条，存isPrime[]。 然后从i=2， 全部变true.
- hash key: the number itself
- 然后利用这个因子的性质，非prime满足条件： self*self, self * self + self ... etc.
- 所以就check每一个j, j+i, j+i+i, 然后把所有non-prime全部mark成false.
- 最后，数一遍还剩下的true个数就好了
```
/*
Description:
Count the number of prime numbers less than a non-negative number, n.
Tags: Hash Table, Math
Similar Problems: (E) Ugly Number, (M) Ugly Number II, (M) Perfect Squares
*/
/*
Attempt2: https://leetcode.com/problems/count-primes/ explains it well
1. Ignore 1 and n. Don't count 1 and the number itself in.
2. Assume all numbers are prime in a boolean[]. Check off those are certainly not prime, the remaining will be prime.
3. For any n, only need to check up to i * i < n; more than that,
for example 2 x 6 is same as checking 6x2, but 6x2 is not necessary to check.
4. How to mark things off:
The first non-prime is always i^2: self * self.
Then more non-primes:self * self, self * (self + 1), self * (self + 2) ... etc.
So, mark all of these index of in the boolean[]
*/
public class Solution {
public int countPrimes(int n) {
if (n <= 1) {
return 0;
}
boolean[] primes = new boolean[n]; // less than n, end prime[n-1]
for (int i = 2; i < primes.length; i++) {
primes[i] = true;
}
for (int i = 2; i * i < n; i++) {
if (!primes[i]) {
continue;
}
for (int j = i * i; j < n; j += i) {
primes[j] = false;
}
}
int count = 0;
for (int i = 2; i < primes.length; i++) {
count += primes[i] ? 1 : 0;
}
return count;
}
}
/*Timeout version*/
//prime is a number n that cannot be divided by any number < n.
//In fact, only need to check sqrt(n) numbers from 1
public class Solution {
public int countPrimes(int n) {
int count = 0;
for (int i = 1; i < n; i++) {
if (isPrime(i)) {
count++;
}
}
return count;
}
public boolean isPrime(int num) {
if (num <= 1) return false;
for (int i = 2; i * i <= num; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
}
```