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 = ; 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.