简单的JAVA四则运算.doc

package arrays.myArray;import java.util.Scanner;public class SortObject private static int intercePosition = 0; / 记录单个运算数据的长度 private static int intercePositionIndex = null; / 记录“（”的下标 private static int intercePositionEnd = null; / 记录“）”的下标 public static void main(String args) Scanner input = new Scanner(System.in); do System.out.println(请输入你要计算的字符串(注意：只能输入数字和加，减，乘除符号;输入完毕后,请直接回车):); String numberString = input.next().trim(); / 判断输入的运算字符串是否符合规定 if (ispassString(numberString) = false) System.out.println(您输入的计算字符串有误,请正确输入!); else / 计算结果返回 System.out.println(interceResult(numberString); while (true); / 判断是否有带括号的运算字符串存在 private static String interceResult(String str) String result = str; char numberString = str.toCharArray(); / 1+2+(1*2+1-1*2+5)+2+(1+5+9+10-11)+1*5/2+3 / 1+8-9+(1*8/2-5+(1+2+8)+4/5*8/3*2 int IndexStart = 0; / 记录“（”的实际数量 int EndStart = 0; / 记录“）”的实际数量 for (int i = 0; i numberString.length; i+) if ( = numberStringi) / 记录最后一个正括号的位置 IndexStart = i; if () = numberStringi) / 记录反括号的最初始下标的位置 EndStart = i; / 截取最里面一个括号里的运算字符串 result = result.substring(IndexStart + 1, EndStart); / 截取括号的运算字符串进行运算,生成新的运算字符串 result = str.substring(0, IndexStart) + interceptOperation(result, *, /) + str.substring(EndStart + 1, str.length(); / 回调执行，其它小括号的运算字符串 return interceResult(result); if (i = numberString.length - 1) if (EndStart = 0) break; / 不存在括号了，再进行混合运算 result = interceptOperation(str, *, /); return result; / 不带括号的四则运算 private static String interceptOperation(String operationNumber, char a, char b) String mess = operationNumber; char stringOperation = mess.toCharArray(); / 循环遍历运算字符串，并做相应的运算 for (int i = 0; i stringOperation.length; i+) / 判断运算符所在的索引 if (stringOperationi = a | stringOperationi = b) if (i != 0) / 运算符前的第一个数 double num1 = interceptNumIndex(mess.substring(0, i); / 记录第一个数据的长度 int frontPosition = intercePosition; / 运算符前的第二个数 double num2 = interceptNumEnd(mess.substring(i + 1, stringOperation.length); / 记录第二个数据的长度 int backPosition = intercePosition; / 算完乘除，将结果替换到原来运算的位置，得到新的运算字符串 String IndexMess = mess.substring(0, i - frontPosition + 1); String IndexResult = ; / 判断是否运算到最后的结果了 if (IndexMess.indexOf(+) = -1 & IndexMess.indexOf(*) = -1 & IndexMess.indexOf(/) = -1 & IndexMess.lastIndexOf(-) = -1) IndexMess = ; if (IndexMess != ) IndexResult = IndexMess.lastIndexOf(-) = IndexMess .length() - 1 ? IndexMess.substring(0, i - frontPosition) : IndexMess; / 组装新的运算字符串 mess = IndexResult/ mess.substring(0,i-frontPosition+1) + reslutString( + stringOperationi, num1, num2) + mess.substring(i + backPosition + 1); / 0.111/1212/2/2/2/2/2/2/2 if (mess.lastIndexOf(-) = 0 & mess.indexOf(+) = -1 & mess.indexOf(*) = -1 & mess.indexOf(/) = -1) break; / 回调，继续运算 return interceptOperation(mess, a, b);/ 1+7-5+89/3+4-6*8/2+4-6 else continue; if (i = stringOperation.length - 1) / 递归出口，判断是否还有运算字符串在 if (mess.indexOf(+) != -1 | mess.indexOf(-) != -1) return interceptOperation(mess, +, -); break; return mess; / 截取第二个数 private static double interceptNumEnd(String str) double a = 0; int InrerceIndex = 0; char stringOperation = str.toCharArray(); boolean ispas = false; / 记录是否为负数 for (int i = 0; i = 0; i-) switch (stringOperationi) case *: case /: case +: case -: InrerceIndex = i; temp = true; break; default: break; if (temp) break; / 判断此数据是否在运算字符串的第一位 if (InrerceIndex = 0) a = Double.parseDouble(str); intercePosition = str.length(); / if(temp) / intercePosition+; else a = Double.parseDouble(str.substring(InrerceIndex, str.length(); / 记录数据的真实长度 intercePosition = str.substring(InrerceIndex, str.length() .length(); return a; / 计算结果 private static double reslutString(String operation, double num1, double num2) double sumResult = 0; if (operation.equals(*) sumResult = num1 * num2; if (operation.equals(-) sumResult = num1 - num2; if (operation.equals(/) sumResult = num1 / num2; if (operation.equals(+) sumResult = num1 + num2; return sumResult; / 判断是否正确输入运算方式 private static boolean ispassString(String messString) boolean ispass = false; boolean operationIspass = true; / 记录被除数的状态 int ai = 0; / 记录是否有运算符号的存在 char IsString = messString.toCharArray(); int num1 = 0; int num2 = 0; for (int i = 0; i IsString.length; i+) / 记录有几对小括号的存在 if ( = IsStringi) num1+; if () = IsStringi) num2+; / 判断除数是否为零 if (/ = IsStringi & IsStringi + 1 = 0) ope