Java编程题讲解.doc

Java编程篇 1. 将一个十进制数（byte类型）转化二进制数，将二进制数前后颠倒，再算出颠倒后其对应的十进制数。 public class MyByte{ private byte b; private int a[] = {0,0,0,0,0,0,0,0}; public MyByte(byte b){ this.b = b; } public void tenToSecond(){ int i=0; do{ a[i] = b%2; b = (byte) (b/2); i++; }while(b != 0); } public byte toTen(){ int temp[] = {0,0,0,0,0,0,0,0},j=0; for(int i=7;i=0;i--){ temp[j]=a[i]; j++; } for(int i=0;i8;i++){ b+=temp[i]*2; } return b; } public static void main(String[] args){ MyByte mb = new MyByte((byte) 7); mb.tenToSecond(); System.out.println(转化后的字符：+mb.toTen()); } } 2.实现简易字符串压缩算法：一个长度最大为128的字符串，由字母a-z或者A-Z组成，将其中连续出现2次以上（含2次）的字母转换为出现次数加字母，以达到压缩目的。 import java.util.*; public class ThreadTestA { void stringZip(char inputStr[], int len, char outputStr[]) { int k = 0; for (int i = 0; i len; i++) { int j = i + 1, n = 1; while (j len) { // 选取i的字符比较i后面的 if (inputStr[i] != inputStr[j]) { //若i与j不等，i向后移一位 if (n 1) { outputStr[k++] = (n + ).charAt(0);//数字转化成对应的数字字符 } outputStr[k++] = inputStr[i]; break; } else { //若i与j相等，j向后移一位，再判断 j++; n++; } i = j - 1; } if (j == len n == 1) outputStr[k] = inputStr[i]; if (j == len n 1) { outputStr[k++] = (n + ).charAt(0); outputStr[k] = inputStr[i]; } } for (int j = 0; j = k; j++) { System.out.print(outputStr[j]); } } public static void main(String[] agrs) { Scanner cin = new Scanner(System.in); String str; while (cin.hasNext()) { str = cin.next(); ThreadTestA t = new ThreadTestA(); int len = str.length(); char outputStr[] = new char[len]; t.stringZip(str.toCharArray(), len, outputStr); } } } 用java写一个字符串匹配“{”，“}”。用栈结构 //StringMatch.java import java.util.Stack; public class StringMatch { private String str; public StringMatch(String str) { this.str=str; } public boolean match(){ StackCharacter st = new StackCharacter(); char ch[]