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java - 可以改进解决方案吗？

转载 作者：塔克拉玛干 更新时间：2023-11-03 05:51:20 27 4
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问题陈述

It is the wedding day of Sanchi, the beautiful princess of Byteland. Her fiance Krishna is planning to gift her an awesome ruby necklace. Krishna has currently b -blue rubies, g -green rubies, r-red rubies and y -yellow rubies. He has to arrange the rubies next to each other in a straight line to make the necklace. But, there are a couple of rules to be followed while making this necklace:

  • A blue ruby should be followed by either a blue ruby or a red ruby
  • A green ruby should be followed by either a green ruby or a yellow ruby
  • A red ruby should be followed by either a green ruby or a yellow ruby
  • A yellow ruby should be followed by either a blue ruby or a red ruby
  • If it is possible, we should always start a necklace with a blue or a red ruby

Can you tell what is the maximum possible length of the necklace that Krishna can make. The length of a necklace is the number of rubies in it.

Input Format

  • The first line contains an integer representing b.
  • The second line contains an integer representing r.
  • The third line contains an integer representing y.
  • The fourth line contains an integer representing g.

Constraints

  • 0 <= b, r, y, g <= 2000
  • At least one of b, r, y, g is greater than 0

Output Format

  • A single integer which is the answer to the problem.

Sample TestCase 1

Input
1
1
1
0
Output
3

我的解决方案:

package ruby;

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;

public class CandidateCode {

public static void main(String[] args) {
    try {
        BufferedReader br = new BufferedReader(new 
InputStreamReader(System.in));
        Integer b = new Integer(br.readLine());
        Integer r = new Integer(br.readLine());
        Integer y = new Integer(br.readLine());
        Integer g = new Integer(br.readLine());

        if (b + r + y + g > 0 && b <= 2000 && b <= 2000 && r <= 2000 && 
g <= 2000 && y <= 2000 && b >= 0 && r >= 0
                && g >= 0 && y >= 0) {

            ArrayList<Character> necklace = new ArrayList<Character>();
            CandidateCode rn = new CandidateCode();
            int length = rn.calculateLengthofNecklace(b, r, y, g, 
necklace);
            System.out.println(length);
        }
    } catch (Exception e) {
    }

}

private int calculateLengthofNecklace(Integer b, Integer r, Integer y, Integer g, ArrayList<Character> necklace) {
    if (anyRemaining(b))
        necklace = addBlue(b, r, y, g, necklace);
    else if (anyRemaining(r))
        necklace = addRed(b, r, y, g, necklace);
    else if (anyRemaining(g))
        necklace = addGreen(b, r, y, g, necklace);
    else if (anyRemaining(y))
        necklace = addYellow(b, r, y, g, necklace);
    // System.out.println("necklace :" + necklace.toString());
    return necklace.size();
}

private ArrayList<Character> addRed(Integer b, Integer r, Integer y, Integer g, ArrayList<Character> necklace) {
    char lastruby = necklace.size() > 0 && necklace.get(necklace.size() - 1) != null
            ? necklace.get(necklace.size() - 1) : 'z';
    if (anyRemaining(r) && (lastruby == 'y' || lastruby == 'b' || lastruby == 'z')) {
        necklace.add('r');
        r = r - 1;
        // System.out.println("in addRed\n b:" + b + " r: " + r + " y: " + y
        // + " g: " + g);
    } else {
        return necklace;
    }
    ArrayList<Character> n1 = new ArrayList<Character>(necklace);
    ArrayList<Character> n2 = new ArrayList<Character>(necklace);
    n1 = addGreen(b, r, y, g, n1);
    n2 = addYellow(b, r, y, g, n2);
    necklace = n1.size() > n2.size() ? n1 : n2;
    return necklace;

}

private ArrayList<Character> addYellow(Integer b, Integer r, Integer y, Integer g, ArrayList<Character> necklace) {
    char lastruby = necklace.size() > 0 && necklace.get(necklace.size() - 1) != null
            ? necklace.get(necklace.size() - 1) : 'z';
    if (anyRemaining(y) && (lastruby == 'r' || lastruby == 'g' || lastruby == 'z')) {
        necklace.add('y');
        y = y - 1;
        // System.out.println("in addYellow\n b:" + b + " r: " + r + " y: "
        // + y + " g: " + g);
    } else {
        return necklace;
    }
    ArrayList<Character> n1 = new ArrayList<Character>(necklace);
    ArrayList<Character> n2 = new ArrayList<Character>(necklace);
    n1 = addBlue(b, r, y, g, n1);
    n2 = addRed(b, r, y, g, n2);
    necklace = n1.size() > n2.size() ? n1 : n2;
    return necklace;
}

private ArrayList<Character> addGreen(Integer b, Integer r, Integer y, Integer g, ArrayList<Character> necklace) {
    char lastruby = necklace.size() > 0 && necklace.get(necklace.size() - 1) != null
            ? necklace.get(necklace.size() - 1) : 'z';
    if (anyRemaining(g) && (lastruby == 'r' || lastruby == 'g' || lastruby == 'z')) {
        necklace.add('g');
        g = g - 1;
        // System.out.println("in addGreen\n b:" + b + " r: " + r + " y: " +
        // y + " g: " + g);
    } else {
        return necklace;
    }
    ArrayList<Character> n1 = new ArrayList<Character>(necklace);
    ArrayList<Character> n2 = new ArrayList<Character>(necklace);
    n1 = addGreen(b, r, y, g, n1);
    n2 = addYellow(b, r, y, g, n2);
    necklace = n1.size() > n2.size() ? n1 : n2;
    return necklace;

}

private ArrayList<Character> addBlue(Integer b, Integer r, Integer y, Integer g, ArrayList<Character> necklace) {
    char lastruby = necklace.size() > 0 && necklace.get(necklace.size() - 1) != null
            ? necklace.get(necklace.size() - 1) : 'z';
    if (anyRemaining(b) && (lastruby == 'y' || lastruby == 'b' || lastruby == 'z')) {
        necklace.add('b');
        b = b - 1;
        // System.out.println("in addBlue\n b:" + b + " r: " + r + " y: " +
        // y + " g: " + g);
    } else {
        return necklace;
    }
    ArrayList<Character> n1 = new ArrayList<Character>(necklace);
    ArrayList<Character> n2 = new ArrayList<Character>(necklace);
    n1 = addBlue(b, r, y, g, n1);
    n2 = addRed(b, r, y, g, n2);
    necklace = n1.size() > n2.size() ? n1 : n2;
    return necklace;

}

private boolean anyRemaining(Integer b) {
    return b > 0;

}

}

这个解决方案可以改进吗？

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