Java实现栈之计算器
Java实现栈来做一个将中缀表达式转化为后缀表达式的程序，中缀表达式更符合我们的主观感受，后缀表达式更适合计算机的运算，下面直接上代码吧：
package Character1;import java.util.Stack;public class Calculter { private static Stack<Object> shu = new Stack<Object>(); private static Stack<Object> fu = new Stack<Object>(); static String oldString = "(1+2)+2*(4-1)+6/3"; public static void main(String[] args) { char[] chars = new char[oldString.length()]; for (int i = 0; i < oldString.length(); i++) { chars[i] = oldString.charAt(i); } for (int i = 0; i < chars.length; i++) { if (isCaoZuoFu(chars[i])) { if (chars[i] == '(') { fu.add(chars[i]); } if (chars[i] == ')') { // System.out.println(fu); while ((char) fu.peek() != '(') { shu.add(fu.pop()); } if ((char) fu.peek() == '(') { fu.pop(); }// System.out.println(fu+"-----"); while (fu.isEmpty() != true) { char temp = (char) fu.pop(); fu.add(temp); if (temp == '*' || temp == '/') { shu.add(temp); fu.pop(); } else { break; } } } else if (chars[i] == '*') { fu.add(chars[i]); } else if (chars[i] == '/') { fu.add(chars[i]); } else if (chars[i] == '+') { fu.add(chars[i]); } else if (chars[i] == '-') { fu.add(chars[i]); } } else { shu.add(chars[i]); } }// System.out.println(fu.toString());// System.out.println(shu.toString()); System.out.println("原字符串是："+oldString); System.out.println("------------------以上部分千万不要改动了"); while(fu.isEmpty()!=true){ shu.add(fu.pop()); } System.out.println("操作符栈："+fu.toString()); System.out.println("操作数栈："+shu.toString()); System.out.println("----------------------中缀表达式转换为后缀表达式的结果！"); int result=JiSuan(Reverse(shu)); System.out.println("表达式最终答案："+result); } public static int JiSuan(Stack<Object> stack){ int result=0; Stack<Object>temp=new Stack<Object>();//存放后缀表达式的操作数的栈 while(stack.isEmpty()!=true){ if(!isCaoZuoFu((char)stack.peek())){ temp.add(stack.pop()); System.out.println(temp); }else{ int o1=Integer.parseInt(temp.pop().toString()); int o2=Integer.parseInt(temp.pop().toString()); if(stack.isEmpty()!=true){ if((char)stack.peek()=='+'){ temp.add(o2+o1); } if((char)stack.peek()=='-'){ temp.add(o2-o1); } if((char)stack.peek()=='*'){ temp.add(o2*o1); } if((char)stack.peek()=='/'){ temp.add(o2/o1); } stack.pop(); } } } return (int) temp.peek(); } public static Stack<Object> Reverse(Stack<Object> shu){ Stack<Object> reverseStack=new Stack<Object>(); while(shu.isEmpty()!=true){ reverseStack.add(shu.pop()); } System.out.println("来个反序："+reverseStack); return reverseStack; } //此方法用于判断对应字符是不是操作符，其中case的数字是对应的ASCII码。 public static boolean isCaoZuoFu(char b) { int temp = (char) b; switch (temp) { case 43: case 45: case 42: case 47: case 40: case 41: return true; default: return false; } }}- 1
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下面就是程序的运行结果：
由于时间匆忙，未来得及做注释，导致此程序可读性较差，还望读者取精华弃糟粕。也希望对我的书写中有错误的地方予以指正，旨在我们的共同进步！
再分享一下我老师大神的人工智能教程吧。零基础！通俗易懂！风趣幽默！还带黄段子！希望你也加入到我们人工智能的队伍中来！https://blog.****.net/jiangjunshow
Java实现栈来做一个将中缀表达式转化为后缀表达式的程序，中缀表达式更符合我们的主观感受，后缀表达式更适合计算机的运算，下面直接上代码吧：
package Character1;import java.util.Stack;public class Calculter { private static Stack<Object> shu = new Stack<Object>(); private static Stack<Object> fu = new Stack<Object>(); static String oldString = "(1+2)+2*(4-1)+6/3"; public static void main(String[] args) { char[] chars = new char[oldString.length()]; for (int i = 0; i < oldString.length(); i++) { chars[i] = oldString.charAt(i); } for (int i = 0; i < chars.length; i++) { if (isCaoZuoFu(chars[i])) { if (chars[i] == '(') { fu.add(chars[i]); } if (chars[i] == ')') { // System.out.println(fu); while ((char) fu.peek() != '(') { shu.add(fu.pop()); } if ((char) fu.peek() == '(') { fu.pop(); }// System.out.println(fu+"-----"); while (fu.isEmpty() != true) { char temp = (char) fu.pop(); fu.add(temp); if (temp == '*' || temp == '/') { shu.add(temp); fu.pop(); } else { break; } } } else if (chars[i] == '*') { fu.add(chars[i]); } else if (chars[i] == '/') { fu.add(chars[i]); } else if (chars[i] == '+') { fu.add(chars[i]); } else if (chars[i] == '-') { fu.add(chars[i]); } } else { shu.add(chars[i]); } }// System.out.println(fu.toString());// System.out.println(shu.toString()); System.out.println("原字符串是："+oldString); System.out.println("------------------以上部分千万不要改动了"); while(fu.isEmpty()!=true){ shu.add(fu.pop()); } System.out.println("操作符栈："+fu.toString()); System.out.println("操作数栈："+shu.toString()); System.out.println("----------------------中缀表达式转换为后缀表达式的结果！"); int result=JiSuan(Reverse(shu)); System.out.println("表达式最终答案："+result); } public static int JiSuan(Stack<Object> stack){ int result=0; Stack<Object>temp=new Stack<Object>();//存放后缀表达式的操作数的栈 while(stack.isEmpty()!=true){ if(!isCaoZuoFu((char)stack.peek())){ temp.add(stack.pop()); System.out.println(temp); }else{ int o1=Integer.parseInt(temp.pop().toString()); int o2=Integer.parseInt(temp.pop().toString()); if(stack.isEmpty()!=true){ if((char)stack.peek()=='+'){ temp.add(o2+o1); } if((char)stack.peek()=='-'){ temp.add(o2-o1); } if((char)stack.peek()=='*'){ temp.add(o2*o1); } if((char)stack.peek()=='/'){ temp.add(o2/o1); } stack.pop(); } } } return (int) temp.peek(); } public static Stack<Object> Reverse(Stack<Object> shu){ Stack<Object> reverseStack=new Stack<Object>(); while(shu.isEmpty()!=true){ reverseStack.add(shu.pop()); } System.out.println("来个反序："+reverseStack); return reverseStack; } //此方法用于判断对应字符是不是操作符，其中case的数字是对应的ASCII码。 public static boolean isCaoZuoFu(char b) { int temp = (char) b; switch (temp) { case 43: case 45: case 42: case 47: case 40: case 41: return true; default: return false; } }}- 1
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- 111
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- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
下面就是程序的运行结果：
由于时间匆忙，未来得及做注释，导致此程序可读性较差，还望读者取精华弃糟粕。也希望对我的书写中有错误的地方予以指正，旨在我们的共同进步！