要想计算24点游戏的结果,则必须要采用基于搜索的算法(即穷举法)对每种情况进行遍历,我们怎么样才能遍历所有的情况呢?其实我们只要总结一下,还是有规律可以找的。
输入a、b、c、d,组成a Op1 bOp2 c Op3 d的表达式,其中先算哪个子表达式未知,一共有5种计算方式,如下图所示:
此时如果要实现该程序,需要存储5棵树,为了能够使得存储量达到最小,通过分析,其实总的来说,只需要存储2棵树即可,即:
其他树都是冗余的,因为我们可以通过a、b、c、d的交换,比如((a+(b*c))+d)可以变为(((b*c)+a)+d);
对于每棵树来说,abcd的可能性为4*3*2*1=24;op1op2 op3的可能性为4*4*4=64,因此总个数为1536,而两棵树的总个数为3072。因此只需要穷举这些方法,就可以知道结果。
TfUtils类为实现穷举24点所有可能情况的类,calculate函数用于计算,参数a、b、c、d分别为给定的4个数,而TfUtils类中的expr属性为求解的表达式。
CalculatorUtils.java
package org.xiazdong; import java.util.Stack; public class CalculatorUtils { /** * 计算后缀表达式 */ public static String calculateReversePolish(String str) { String[] splitStr = str.split(" "); Stack<String> s = new Stack<String>(); for (int i = 0; i < splitStr.length; i++) { String ch = splitStr[i]; if (ch.matches("\\d+.\\d+")||ch.matches("\\d+")) { s.push(ch); } else { if (s.size() >= 2) { String c1 = s.pop(); String c2 = s.pop(); if (ch.equals("+")) { if(c1.contains(".")||c2.contains(".")){ s.push(String.valueOf((Double.parseDouble(c2 + "") + Double .parseDouble(c1 + "")))); } else{ s.push(String.valueOf((Integer.parseInt(c2 + "") + Integer .parseInt(c1 + "")))); } } else if ("-".equals(ch)) { if(c1.contains(".")||c2.contains(".")){ s.push(String.valueOf((Double.parseDouble(c2 + "") - Double .parseDouble(c1 + "")))); } else{ s.push(String.valueOf((Integer.parseInt(c2 + "") - Integer .parseInt(c1 + "")))); } } else if ("*".equals(ch)) { if(c1.contains(".")||c2.contains(".")){ s.push(String.valueOf((Double.parseDouble(c2 + "") * Double .parseDouble(c1 + "")))); } else{ s.push(String.valueOf((Integer.parseInt(c2 + "") * Integer .parseInt(c1 + "")))); } } else if ("/".equals(ch)) { s.push(String.valueOf((Double.parseDouble(c2 + "") / Double .parseDouble(c1 + "")))); } } else { System.out.println("式子有问题!"); return null; } } } return s.pop(); } }
package org.xiazdong; import java.io.Serializable; public class TfUtils implements Serializable{ private int result; private String expr = ""; //存放中缀表达式 public String getExpr() { return expr; } public void setExpr(String expr) { this.expr = expr; } /* (操作符1) / \ (操作符2) (操作数4) / \ (操作符3) (操作数3) / \ (操作数1) (操作数2) */ private int tree1[] = new int[7];; // 存放第一棵树 //private int tree2[]; // 存放第二棵树 private final int PLUS = 1; // 加 private final int MINUS = 2; // 减 private final int MULT = 3; // 乘 private final int DIV = 4; // 除 /** * 计算24点的主函数 */ public void calculate(int a, int b, int c, int d) { int data[] = { a, b, c, d }; // 1.用数组构建一棵树,其中0,1,3处填操作符;2,4,5,6填充操作数 // 2.按照参数a,b,c,d不同顺序填充树,+-*/也填充 for (int h = 0; h < 4; h++) { for (int i = 0; i < 4; i++) { if (i == h) { continue; } for (int j = 0; j < 4; j++) { if (j == i || j == h) { continue; } for (int k = 0; k < 4; k++) { if (k == h || k == i || k == j) { continue; } tree1[2] = data[h]; tree1[4] = data[i]; tree1[5] = data[j]; tree1[6] = data[k]; // 填充操作符 for (int m = PLUS; m <= DIV; m++) { for (int n = PLUS; n <= DIV; n++) { for (int o = PLUS; o <= DIV; o++) { tree1[0] = m; tree1[1] = n; tree1[3] = o; String t[] = new String[4]; for (int z = 0; z < 4; z++) { switch (tree1[z]) { case PLUS: t[z] = "+"; break; case MINUS: t[z] = "-"; break; case MULT: t[z] = "*"; break; case DIV: t[z] = "/"; break; } } // 目前为止tree数组全部已赋值 String postexpr = tree1[5] + " " + tree1[6] + " " + t[3] + " " + tree1[4] + " " + t[1] + " " + tree1[2] + " " + t[0]; String result = CalculatorUtils .calculateReversePolish(postexpr); if (Double.parseDouble((result)) == 24.0) { expr = "(((" + tree1[5] + t[3] + tree1[6] + ")" + t[1] + tree1[4] + ")" + t[0] + tree1[2] + ")"; System.out.println(expr); return; } } } } } } } } //tree2 = new int[7]; for (int h = 0; h < 4; h++) { for (int i = 0; i < 4; i++) { if (i == h) { continue; } for (int j = 0; j < 4; j++) { if (j == i || j == h) { continue; } for (int k = 0; k < 4; k++) { if (k == h || k == i || k == j) { continue; } tree1[3] = data[h]; tree1[4] = data[i]; tree1[5] = data[j]; tree1[6] = data[k]; // 填充操作符 for (int m = PLUS; m <= DIV; m++) { for (int n = PLUS; n <= DIV; n++) { for (int o = PLUS; o <= DIV; o++) { tree1[0] = m; tree1[1] = n; tree1[2] = o; String t[] = new String[3]; for (int z = 0; z < 3; z++) { switch (tree1[z]) { case PLUS: t[z] = "+"; break; case MINUS: t[z] = "-"; break; case MULT: t[z] = "*"; break; case DIV: t[z] = "/"; break; } } // 目前为止tree数组全部已赋值 String postexpr = tree1[4] + " " + tree1[3] + " " + t[1] + " " + tree1[6] + " " + tree1[5] + " " + t[2] + " " + t[0]; String result = CalculatorUtils .calculateReversePolish(postexpr); if (Double.parseDouble((result)) == 24.0) { expr = "((" + tree1[3] + t[1] + tree1[4] + ")" + t[0] +"("+tree1[5] + t[2] + tree1[6] + "))"; System.out.println(expr); return; } } } } } } } } expr = "无解"; } public int getResult() { return result; } public void setResult(int result) { this.result = result; } }
TfUtils tf = new TfUtils(); tf.calculate(d1, d2, d3, d4); System.out.println(tf.getExpr());
输出为:(((3/7)+3)*7)