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2013
11-09

POJ 1056 IMMEDIATE DECODABILITY [解题报告] Java

IMMEDIATE DECODABILITY

问题描述 :

An encoding of a set of symbols is said to be immediately decodable if no code for one symbol is the prefix of a code for another symbol. We will assume for this problem that all codes are in binary, that no two codes within a set of codes are the same, that each code has at least one bit and no more than ten bits, and that each set has at least two codes and no more than eight.

Examples: Assume an alphabet that has symbols {A, B, C, D}

The following code is immediately decodable:

A:01 B:10 C:0010 D:0000

but this one is not:

A:01 B:10 C:010 D:0000 (Note that A is a prefix of C)

输入:

Write a program that accepts as input a series of groups of records from standard input. Each record in a group contains a collection of zeroes and ones representing a binary code for a different symbol. Each group is followed by a single separator record containing a single 9; the separator records are not part of the group. Each group is independent of other groups; the codes in one group are not related to codes in any other group (that is, each group is to be processed independently).

输出:

For each group, your program should determine whether the codes in that group are immediately decodable, and should print a single output line giving the group number and stating whether the group is, or is not, immediately decodable.

样例输入:

01
10
0010
0000
9
01
10
010
0000
9

样例输出:

Set 1 is immediately decodable
Set 2 is not immediately decodable
 

解题代码:

//* @author 娲��楣�lt;[email protected]>

import java.util.ArrayList;
import java.util.Scanner;

public class Main {

/**
 * @param args
 */
 public static boolean isPrefix(String a, String b) {
	int len = a.length() > b.length() ? b.length() : a.length();
	if (a.substring(0, len).equals(b.substring(0, len)))
      return true;
	else
		return false;
 }

public static void main(String[] args) {
  ArrayList< String> array = new ArrayList< String>();
  Scanner in = new Scanner(System.in);
  int num = 1;
  boolean skip = false;
  while (in.hasNext()) {
	String temp = in.next();
	if (temp.equals("9")) {
	 int len = array.size();
	 for (int i = 0; i < len - 1; i++) {
	  for (int j = i + 1; j < len; j++) {
		if (isPrefix(array.get(i), array.get(j))) {
		 skip = true;
		}
	   }
	}
	if (!skip) {
	 System.out.println("Set " + num+ " is immediately decodable");
	} else {
	 System.out.println("Set " + num+ " is not immediately decodable");
	}
		skip = false;
		num++;
		array.clear();
		continue;
	} else {
		array.add(temp);
	}
  }
 }

}

  1. int half(int *array,int len,int key)
    {
    int l=0,r=len;
    while(l<r)
    {
    int m=(l+r)>>1;
    if(key>array )l=m+1;
    else if(key<array )r=m;
    else return m;
    }
    return -1;
    }
    这种就能避免一些Bug
    l,m,r
    左边是l,m;右边就是m+1,r;

  2. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。