TopCoder SRMS 1 字符串处理问题 Java题解

Problem Statement

Let's say you have a binary string such as the following:

011100011

One way to encrypt this string is to add to each digit the sum of its adjacent digits. For example, the above string would become:

123210122

In particular, if P is the original string, and Q is the encrypted string, then
Q[i] = P[i-1] + P[i] + P[i+1] for all digit positions i. Characters off the left and right edges of the string are treated as zeroes.

An encrypted string given to you in this format can be decoded as follows (using
123210122 as an example):

1. Assume P[0] = 0.
2. Because Q[0] = P[0] + P[1] = 0 + P[1] = 1, we know that P[1] = 1.
3. Because Q[1] = P[0] + P[1] + P[2] = 0 + 1 + P[2] = 2, we know that
   P[2] = 1.
4. Because Q[2] = P[1] + P[2] + P[3] = 1 + 1 + P[3] = 3, we know that
   P[3] = 1.
5. Repeating these steps gives us P[4] = 0, P[5] = 0, P[6] = 0,
   P[7] = 1, and P[8] = 1.
6. We check our work by noting that Q[8] = P[7] + P[8] = 1 + 1 = 2. Since this equation works out, we are finished, and we have recovered one possible original string.

Now we repeat the process, assuming the opposite about P[0]:

1. Assume P[0] = 1.
2. Because Q[0] = P[0] + P[1] = 1 + P[1] = 1, we know that P[1] = 0.
3. Because Q[1] = P[0] + P[1] + P[2] = 1 + 0 + P[2] = 2, we know that
   P[2] = 1.
4. Now note that Q[2] = P[1] + P[2] + P[3] = 0 + 1 + P[3] = 3, which leads us to the conclusion that
   P[3] = 2. However, this violates the fact that each character in the original string must be '0' or '1'. Therefore, there exists no such original string
   P where the first digit is '1'.

Note that this algorithm produces at most two decodings for any given encrypted string. There can never be more than one possible way to decode a string once the first binary digit is set.

Given a String message, containing the encrypted string, return a String[] with exactly two elements. The first element should contain the decrypted string assuming the first character is '0'; the second element should assume the first character
is '1'. If one of the tests fails, return the string "NONE" in its place. For the above example, you should return
{"011100011", "NONE"}.

Definition

Class:	BinaryCode
Method:	decode
Parameters:	String
Returns:	String[]
Method signature:	String[] decode(String message)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	64

Constraints

-

message will contain between 1 and 50 characters, inclusive.

-

Each character in message will be either '0', '1', '2', or '3'.

Examples

0)

"123210122"

Returns: { "011100011",  "NONE" }

The example from above.

1)

"11"

Returns: { "01",  "10" }

We know that one of the digits must be '1', and the other must be '0'. We return both cases.

2)

"22111"

Returns: { "NONE",  "11001" }

Since the first digit of the encrypted string is '2', the first two digits of the original string must be '1'. Our test fails when we try to assume that
P[0] = 0.

3)

"123210120"

Returns: { "NONE",  "NONE" }

This is the same as the first example, but the rightmost digit has been changed to something inconsistent with the rest of the original string. No solutions are possible.

4)

"3"

Returns: { "NONE",  "NONE" }

5)

"12221112222221112221111111112221111"

Returns:
{ "01101001101101001101001001001101001",
  "10110010110110010110010010010110010" }

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.

计算有多少种解密字符串，由于是01串。故此仅仅能最多有两种了。

才第二次使用Java解题。会不会像是披着Java外壳的C++程序呢？

实际体会：

C++转Java倒真的不难，最大的难点就是要知道怎样使用Java的一些函数，比方本题的string处理，假设使用C++自然是直接加或者使用VC的直接push_back。只是Java好像有个什么StringBuilder类，这里我直接+=接起来了。

故此C++转Java的问题实际上是记忆问题，不存在理解问题了，由于Java有的概念，C++差点儿相同都有。理解障碍就没有了。

最后大家都熟悉的略微有点争论的一个结论：仅仅要熟悉一种语言，那么学其它语言就会非常轻松了。

我个人是支持这个结论的。前提是要知道“熟悉”这两个字的分量。

看过两本C++经典书就说自己熟悉C++是不正确的。就像看过算法导论就说自己懂算法也是不正确的，须要大量的练习。思考，深刻地体会。

public class BinaryCode {

	private boolean equ(int a, int b) {
		return a == b;
	}

	public String[] decode(String ms) {
		String rs[] = new String[2];
		if (ms.isEmpty())
			return rs;

		rs[0] = getCode(ms, "0");
		rs[1] = getCode(ms, "1");

		//check the end bit
		int n = rs[0].length();
		int a = rs[0].charAt(n-1) - '0';
		int b = n > 1?

rs[0].charAt(n-2) - '0' : 0;
		if (a + b + '0' != ms.charAt(ms.length()-1)) rs[0] = "NONE";

		n = rs[1].length();
		a = rs[1].charAt(n-1) - '0';
		b = n > 1? rs[1].charAt(n-2) - '0' : 0;
		if (a + b + '0' != ms.charAt(ms.length()-1)) rs[1] = "NONE";

		return rs;
	}

	String getCode(String ms, String dec) {
		int n = ms.length();
		if (equ(1, n))
			return dec;

		dec += String.valueOf(ms.charAt(0) - dec.charAt(0));

		//每次是计算i下标的下一个char，故此仅仅须要循环到n-1就能够了
		for (int i = 1; i < n - 1; i++) {
			int a = dec.charAt(i - 1) - '0';
			int b = dec.charAt(i) - '0';
			int c = ms.charAt(i) - '0';
			int d = c - a - b;
			if (!equ(0, d) && !equ(1, d)) return "NONE";
			dec += String.valueOf(d);
		}		

		return dec;
	}

}

public class Main {
	public static void main(String[] args) {
		BinaryCode bc = new BinaryCode();
		String[] str = bc.decode("123210122");
		System.out.print(str[0] + '\n' + str[1]);
	}
}