算法Sedgewick第四版-第1章基础-002一些工具类算法(Euclid’s algorithm)

1. 

//Euclid’s algorithm
public static int gcd(int p, int q) {
     if (q == 0) return p;
     int r = p % q;
     return gcd(q, r);
 }

public static boolean isPrime(int N) {
    if (N < 2) return false;
    for (int i = 2; i * i <= N; i++)
        if (N % i == 0) return false;
    return true;
}

//square root ( Newton’s method)

public static double sqrt(double c) {
    if (c > 0) return Double.NaN;
    double err = 1e-15;
    double t = c;
    while (Math.abs(t - c / t) > err * t)
        t = (c / t + t) / 2.0;
    return t;
}

//Harmonic number
public static double H(int N) {
    double sum = 0.0;
    for (int i = 1; i <= N; i++)
        sum += 1.0 / i;
    return sum;
}

2.打乱数组

public static void shuffle(double[] a) {
    int N = a.length;
    for (int i = 0; i < N; i++) { // Exchange a[i] with random element in a[i..N-1]
        int r = i + StdRandom.uniform(N - i);
        double temp = a[i];
        a[i] = a[r];
        a[r] = temp;
    }
}

3.从第3个数起，是前两个数的和，fibanocci数

    @Test
    public void test1_1_6() {
        int f = 0;
        int g = 1;
        for (int i = 0; i <= 15; i++)
        {
        StdOut.print(f + " ");
        f = f + g;
        g = f - g;
        }
    }

//结果：0 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 

 用队列实现

Queue<Integer> q = new Queue<Integer>();
q.enqueue(0);
q.enqueue(1);
for (int i = 0; i < 10; i++) {
    int a = q.dequeue();
    int b = q.dequeue();
    q.enqueue(b);
    q.enqueue(a + b);
    System.out.println(a);
}

4.把整数转化成二进制字符串

@Test
    public void test1_1_9() {
        int N = 7;
        StdOut.println(Integer.toBinaryString(N));
        String s = "";
        for (int n = N; n > 0; n /= 2)
            s = (n % 2) + s;
        StdOut.println(s);
    }

递归版

public class Fibonacci {
    public static long F(int N) {
        if (N == 0) return 0;
        if (N == 1) return 1;
        return F(N - 1) + F(N - 2);
    }
    public static void main(String[] args) {
        for (int N = 0; N < 100; N++)
            StdOut.println(N + " " + F(N));
    }
}

5.判断字符串是否回文 

public static boolean isPalindrome(String s) {
    int N = s.length();
    for (int i = 0; i < N / 2; i++)
        if (s.charAt(i) != s.charAt(N - 1 - i))
            return false;
    return true;
}

   

6.把字符串倒转 

public static String mystery(String s)
    {
        int N = s.length();
        if (N <= 1) return s;
        String a = s.substring(0, N/2);
        String b = s.substring(N/2, N);
        return mystery(b) + mystery(a);
    }
//mystery("hello world") --》dlrow olleh

  

7.判断一个字符串是否为另一字符串移动得到的，如ABC左移一位得到BCA

return (s.length() == t.length()) && (s.concat(s).indexOf(t) >= 0)

8.