[LeetCode] 204. count primes

Topic links: https://leetcode-cn.com/problems/count-primes/

Subject description:

All statistical non-negative integers less than n number of prime numbers.

Example:

输入: 10
输出: 4
解释: 小于 10 的质数一共有 4 个, 它们是 2, 3, 5, 7 。

Ideas:

Prime number is in addition to 1and in itself can not find other number divisible, see the topic ideas tips!

A thought: Violence Act (timeout) (you can learn about for ... elseusage, with the general breakuse)

class Solution:
    def countPrimes(self, n: int) -> int:
        res = 0
        for i in range(2, n):
            for j in range(2, i):
                if i % j == 0:
                    break
            else:
                #print(i)
                res += 1
        return res

Ideas II: optimization of violence (timeout), we can not require verification prime number less than its number are verified

class Solution:
    def countPrimes(self, n: int) -> int:
        res = 0
        for i in range(2, n):
            for j in range(2, int(i ** 0.5) + 1):
                if i % j == 0:
                    break
            else:
                # print(i)
                res += 1
        return res

Thinking three: Eladuose sieve (seems able to live)

class Solution:
    def countPrimes(self, n: int) -> int:
        isPrimes = [1] * n
        res = 0
        for i in range(2, n):
            if isPrimes[i] == 1: res += 1
            j = i
            while i * j < n:
                isPrimes[i * j] = 0
                j += 1
        return res

Ideas Four: Comprehensive Optimization (over 90%) with the

class Solution:
    def countPrimes(self, n: int) -> int:
        if n < 2: return 0
        isPrimes = [1] * n
        isPrimes[0] = isPrimes[1] = 0
        for i in range(2, int(n ** 0.5) + 1):
            if isPrimes[i] == 1:
                isPrimes[i * i: n: i] = [0] * len(isPrimes[i * i: n: i])
        return sum(isPrimes)