算法(JAVA)----两道小小算法题

1、编写一个程序解决选择问题。令k=N/2,画出表格显示程序对于N种不同的值的运行时间。

 

         分析:选择问题是指从N个数当中,按升序(降序也可以)排列,找出第k个数。LZ的写法是采用书中给出的算法自己实现的,分别采用冒泡排序和分批处理的方式。以下为LZ写出的算法代码。

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import java.util.Arrays;
import java.util.Random;

//选择问题答案
public class Select {
    
    public static final Random RANDOM = new Random(47);
    
    //假设N = 10
    public static void main(String[] args) {
        for (int i = 0; i < 10; i++) {
            printResult(createArray(RANDOM.nextInt(100000)));
        }
    }
    
    //冒泡排序
    public static void sort(int[] values){
        for (int i = 1; i < values.length; i++) {
            for (int j = 0; j < values.length - i; j++) {
                if (values[j] > values[j + 1]) {
                    int temp = values[j];
                    values[j] = values[j + 1];
                    values[j + 1] = temp;
                }
            }
        }
    }
    //分批处理
    public static int select(int[] values){
        if (values == null || values.length == 0) {
            throw new NullPointerException("values can't be null.");
        }
        int k = values.length / 2;
        int[] temp = Arrays.copyOf(values, k);
        sort(temp);
        for (int i = k ; i < values.length; i++) {
            if (values[i] < temp[k - 1]) {
                temp[k - 1] = temp[k - 2];
                for (int j = k - 2; j >0; j--) {
                    if (values[i] > temp[j]) {
                        temp[j + 1] = values[i];
                        break;
                    }else {
                        temp[j] = temp[j - 1];
                    }
                }
            }
        }
        return temp[k - 1];
    }
    //创建随即数组
    public static int[] createArray(int length){
        int[] values = new int[length];
        for (int i = 0; i < length; i++) {
            values[i] = RANDOM.nextInt(length * 2);
        }
        return values;
    }
    //打印结果
    public static void printResult(int[] values){
        System.out.println("length:" + values.length);
        long start = System.currentTimeMillis();
        System.out.println("result:" + select(values));
        System.out.println("cost:" + (System.currentTimeMillis() - start) + "ms");
        System.out.println("--------------------------------");
    }
    
}
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2、编写一个程序求解字谜游戏问题。

 

          分析:字谜游戏是从一个二维的字符数组中按照标量寻找单词的过程,LZ写的算法就是最直观的算法,有很多嵌套循环,倘若字符表很大的时候,这种方式会很慢。想必书中的后面章节应该会有更简单的方式,只不过LZ还没看到,以后有机会改善这个算法。

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import java.util.Random;

//字谜问题
public class Character {
    
    public static final Random RANDOM = new Random(47);
    
    public static final String[] WORDS = new String[]{"ynz","yzgm","oqz","owznt","z"};
    
    public static void main(String[] args) {
        char[][] chars = createTable(5);
        printTable(chars);
        findWord(chars);
    }
    //按照标量方向寻找满足的单词(或者说字符串)
    public static void findWord(char[][] chars){
        long start = System.currentTimeMillis();
        for (int i = 0; i < chars.length; i++) {
            for (int j = 0; j < chars.length; j++) {
                for (int k = 0; k < chars.length; k++) {
                    for (int l = 0; l < chars.length; l++) {
                        if (i == k && j == l) {
                            printWord(String.valueOf(chars[i][j]), i, j, k, l);
                            continue;
                        }
                        if (k != i && j != l && (k - i) != (j - l) && (k - i) != (l - j)) {
                            continue;
                        }
                        StringBuffer stringBuffer = new StringBuffer();
                        if (i == k) {
                            if (j > l) {
                                for (int m = j; m >= l; m--) {
                                    stringBuffer.append(chars[i][m]);
                                }
                            }else {
                                for (int m = j; m <= l; m++) {
                                    stringBuffer.append(chars[i][m]);
                                }
                            }
                        }
                        if (j == l) {
                            if (i > k) {
                                for (int m = i; m >= k; m--) {
                                    stringBuffer.append(chars[m][j]);
                                }
                            }else {
                                for (int m = i; m <= k; m++) {
                                    stringBuffer.append(chars[m][j]);
                                }
                            }
                        }
                        if ((k - i) == (j - l)) {
                            if (i > k) {
                                for (int m = i,n = j; m >= k && n <= l; m--,n++) {
                                    stringBuffer.append(chars[m][n]);
                                }
                            }else {
                                for (int m = i,n = j; m <= k && n >= l; m++,n--) {
                                    stringBuffer.append(chars[m][n]);
                                }
                            }
                        }
                        if ((k - i) == (l - j)) {
                            if (i > k) {
                                for (int m = i,n = j; m >= k && n >= l; m--,n--) {
                                    stringBuffer.append(chars[m][n]);
                                }
                            }else {
                                for (int m = i,n = j; m <= k && n <= l; m++,n++) {
                                    stringBuffer.append(chars[m][n]);
                                }
                            }
                        }
                        printWord(stringBuffer.toString(), i, j, k, l);
                    }
                }
            }
        }
        System.out.println("-------------------------------------------------");
        System.out.println("cost time:" + (System.currentTimeMillis() - start) + "ms");
    }
    //判断是否是既定的一个单词(或字符串)并打印
    public static void printWord(String word,int i ,int j,int k,int l){
        for (int m = 0; m < WORDS.length; m++) {
            if (word.equals(WORDS[m])) {
                System.out.println("find word:" + WORDS[m]);
                System.out.println("scalar:" + "[" + (i+1) + "," + (j+1) + "]->[" + (k+1) + "," + (l+1) + "]");
            }
        }
    }
    //创建随即字符二维数组
    public static char[][] createTable(int length){
        char[][] chars = new char[length][length];
        for (int i = 0; i < chars.length; i++) {
            for (int j = 0; j < chars[i].length; j++) {
                chars[i][j] = (char)(97 + RANDOM.nextInt(26));
            }
        }
        return chars;
    }
    //打印二维数组
    public static void printTable(char[][] chars){
        System.out.println("---------------------------------------------");
        for (int i = 0; i < chars.length; i++) {
            System.out.print("	" + (i+1));
        }
        System.out.println();
        for (int i = 0; i < chars.length; i++) {
            System.out.print((i+1));
            for (int j = 0; j < chars.length; j++) {
                System.out.print("	" + chars[i][j]);
            }
            System.out.println();
        }
        System.out.println("---------------------------------------------");
    }
    
}
原文地址:https://www.cnblogs.com/davidshi/p/3349139.html