Algorithms 4th - 1.1 Basic Programming Model - CREATIVE PROBLEMS-优快云博客

　　欢迎交流

　　1.1.26

public class TestApp {
    public static void main(String[] args) {

        int a = StdIn.readInt();
        int b = StdIn.readInt();
        int c = StdIn.readInt();
        int t;

        if( a > b) {
            t = a;
            a = b;
            b = t;
        }

        if( a > c) {
            t = a;
            a = c;
            c = t;
        }

        if( b > c) {
            t = b;
            b = c;
            c = t;
        }

        StdOut.println(a + "->" + b + "->" + c);
    }
}

　　 1.1.27

public class Binomial {
    /**
     * 递归方式的二项分布
     * @param N 总次数
     * @param k 出现次数
     * @param p 每次出现概率
     * @return
     */
    public static double binomial1(int N, int k, double p) {
        if(N == 0 && k == 0)
            return 1.0;
        if(N < 0 || k < 0)
            return 0.0;
        return (1.0 - p) * binomial1(N - 1, k, p) + p * binomial1(N - 1, k - 1, p);
    }

    public static double binomial2(int N, int k, double p) {
        double[][] b = new double[N + 1][N + 1];
        // base
        for(int i = 0; i <= N; i++) {
            b[i][0] = Math.pow(1.0 - p, i);
        }
        b[0][0] = 1.0;
        // recursive
        for(int i = 1; i <= N; i++) {
            for(int j = 1; j <=k; j++) {
                b[i][j] =  p * b[i - 1][j - 1] + (1.0 - p) * b[i - 1][j];
            }
        }
        return b[N][k];
    }

    public static void main(String[] args) {
        int N = 100;
        int k = 50;
        double p = 0.76;
        StdOut.println(binomial1(N, k, p));
        StdOut.println(binomial2(N, k, p));
    }
}

　　1.1.28

　　想不出与BinarySearch有关的去重方法。

　　1.1.29

// returns the number of elements that are smaller than the key
    public static int rank(int key, int[] a) {
        int lo = 0;
        int hi = a.length - 1;
        while(lo <= hi) {
            int mid = (lo + hi) / 2;
            if(key < a[mid])
                hi = mid - 1;
            else if(key > a[mid])
                lo = mid + 1;
            else {
                while(a[mid - 1] == key) {
                    mid -= 1;
                }
                return mid;
            }
        }
        return -1;
    }

    // returns the number of elements equal to the key
    public static int count(int key, int[] a) {
        int lo = 0;
        int hi = a.length - 1;
        while(lo <= hi) {
            int mid = (lo + hi) / 2;
            if(key < a[mid])
                hi = mid - 1;
            else if(key > a[mid])
                lo = mid + 1;
            else {
                lo = mid;
                hi = mid;
                while(a[lo - 1] == key) {
                    lo -= 1;
                }
                while(a[hi + 1] == key) {
                    hi += 1;
                }
                return hi - lo + 1;
            }
        }
        return 0;
    }