Count prime
1. Introduction
Count the number of all prime numbers less than a non-negative integer n
(source of the question: LeetCode )
示例 1：
输入：n = 10
输出：4
解释：小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。
示例 2：
输入：n = 0
输出：0
示例 3：
输入：n = 1
输出：0
提示：
0 <= n <= 5 * 106
Two, the solution
1. Violence Law
class Solution {
public int countPrimes(int n) {
int ans = 0;
//遍历小于n且大于等于2的所有数字
for (int i = 2; i < n; ++i) {
ans += isPrime(i) ? 1 : 0;
}
return ans;
}
//判断这个数字是否为质数
public boolean isPrime(int x) {
for (int i = 2; i * i <= x; ++i) {
if (x % i == 0) {
return false;
}
}
return true;
}
}
2.
The Wikipedia animation:
class Solution {
public int countPrimes(int n) {
//新建一个长度为n的数组(1标记是质数，0代表不是质数)
int[] primes=new int[n];
//初始化全部为质数
Arrays.fill(primes,1);
int count=0;
for (int i = 2; i < n; i++) {
if(primes[i]==1){
//如果当前是质数,计数count++
count++;
//再从i*i开始遍历到n为止(不包含n)，如果是i的倍数肯定不是质数，把它对应的下标赋值为0
if((long)i*i<n){
for (int j=i*i;j<n;j+=i){
primes[j]=0;
}
}
}
}
return count;
}
}
3. Elbow Odd Sieve
As shown in the figure above, we can see that even numbers account for half, and half of the time can be saved by removing them in advance (even numbers must not be prime numbers, except for 2)
class Solution {
public int countPrimes(int n) {
//新建一个长度为n的数组(1标记是质数，0代表不是质数)
int[] primes=new int[n];
//初始化全部为质数
Arrays.fill(primes,1);
//判断当前是否超过了2(超过了2就要统计2这个质数)
int count=n>2?1:0;
for (int i = 3; i < n; i+=2) {
//每次递增改成只遍历奇数
if(primes[i]==1){
count++;
if((long)i*i<n){
for (int j=i*i;j<n;j+=i){
primes[j]=0;
}
}
}
}
return count;
}
}