Count the number of prime numbers less than a non-negative number, n.

Example:

Input: 10
Output: 4
Explanation: There are 4 prime numbers less than 10, they are 2, 3, 5, 7.

//想先用粗暴的方法實現一下
public static int countPrimes(int n) {

    int count = 0;
    for(int i = 1;i<n;i++){
        if(isPrime(i)){
           count ++;
        }
    }
    return count;

}

//全部判斷一下是否書素數 但是這看到有是那個迴圈 時間複雜度接受不了
private static boolean isPrime(int i) {

    if(i ==1){
        return false;
    }
    for(int j =2;j< i;j++){
        for(int j2 =2;j2< i;j2++){
             if(j*j2 == i){
                 return false;
             }
        }
    }
    return true;

}

//如果稍微優化一下呢 結果還是超時
public static boolean isPrime2(int i) {

    if(i ==1 ){
        return false;
    }
    //這樣是否可以快點去掉一些資料
    if(i > 10 && (i%2==0 ||i%3==0||i%5==0||i%7==0)){
        return false;
    }
    for(int j =2;j< Math.sqrt(i)+1;j++){
        int j2 = (int)(Math.sqrt(i)) > 2?  (int)(Math.sqrt(i)):2;
        for(; j2< i; j2++){
            if(j*j2 == i){
                return false;
            }
        }
    }
    return true;

}

/**
  這裡的想法是反一下 找出所有的合數
 */
public static int countPrimes2(int n) {

    boolean[] notPrime = new boolean[n];
    int count = 0;
    for (int i = 2; i < n; i++) {
        if (notPrime[i] == false) {
            count++;
            for (int j = 2; i*j < n; j++) {
                notPrime[i*j] = true;
            }
        }
    }

    return count;
}


//這裡用了不斷累加的方法 感覺也不錯
public static  int countPrimes3(int n) {

    boolean[] m = new boolean[n];
    int count = 0;
    for (int i=2; i<n; i++) {
        if (m[i]) {
            continue;
        }
        count++;
        for (int j=i; j<n; j=j+i) {
            m[j] = true;
        }
    }

    return count;

}

git：https://github.com/woshiyexinjie/leetcode-xin/tree/master/src/main/java/com/helloxin/leetcode/algorithms