逆波兰法（计算器）程序<无括号版>

涉及队列、栈的运用。

Java中队列可以用：

Queue<String> q = new LinkedList();

来声明，其主要的方法有：

poll(),peak(),offer(),clear(),size()等。

 

Java中栈可以用:

Stack s = new Stack();

来声明，其主要方法有：
push(),peak(),pop(),clear(),size()等。

 

  1 package preTest;
  2 
  3 import java.util.LinkedList;
  4 import java.util.Queue;
  5 import java.util.Scanner;
  6 import java.util.Stack;
  7 
  8 //逆波兰表示法
  9 public class nbl {
 10     public static Stack sigStack = new Stack();
 11     public static Queue<String> nblQueue = new LinkedList();
 12     public static Stack numFromQueue = new Stack();
 13     public static void main(String[] args) {
 14         Scanner in=new Scanner(System.in);
 15         while(in.hasNext()){
 16             String str = in.nextLine();
 17             int result = cal(str);
 18             System.out.println(result);
 19         }
 20     }
 21     
 22     public static int cal(String str){
 23         sigStack.clear();
 24         nblQueue.clear();
 25         StringBuilder sb = new StringBuilder();
 26         int len = str.length();
 27         int result= 0;
 28         //压栈和入队操作
 29         for(int i=0; i<len;i++){
 30             char c = str.charAt(i);
 31             if(c>='0' && c <='9'){
 32                 sb.append(c);
 33                 if(i==len-1){
 34                     nblQueue.offer(sb.toString());
 35                 }
 36             }else{
 37                 nblQueue.offer(sb.toString());
 38                 sb.replace(0, sb.length(), "");
 39                 if(sigStack.size() == 0){
 40                     sigStack.push(c);
 41                 }else{
 42                     char existed = (char)sigStack.peek();
 43                     //如果已经压栈的符号优先级>=当前符号c，则完成压栈符号的操作
 44                     if(c=='+' || c=='-'){
 45                         zhengli();
 46                         sigStack.push(c);
 47                         //当两个都是乘除运算时
 48                     }else if(existed =='*' || existed == '/'){
 49                         tinyAdjust(c);
 50                     }else{//新符号乘除，旧符号加减
 51                         sigStack.push(c);
 52                     }    
 53                 }
 54             }
 55         }
 56         zhengli();
 57         //出队操作
 58         numFromQueue.clear();
 59         result = releaseQueue();
 60         return result;
 61     }
 62     
 63     public static void zhengli(){
 64         while(sigStack.size() > 0){
 65             char c = (char)sigStack.pop();
 66             nblQueue.offer(new StringBuilder().append(c).toString());
 67         }
 68     }
 69     
 70     public static void tinyAdjust(char c){
 71         char ch = (char)sigStack.pop();
 72         nblQueue.offer(new StringBuilder().append(ch).toString());
 73         sigStack.push(c);
 74     }
 75     
 76     public static int releaseQueue(){
 77         int len = nblQueue.size();
 78         boolean isNum = false;
 79         for(int i=0; i<len; i++){
 80             isNum = judge();
 81             if(isNum){
 82                 int num = Integer.parseInt(nblQueue.poll());
 83                 numFromQueue.push(num);
 84             }else{
 85                 int a = (int)numFromQueue.pop();
 86                 int b = (int)numFromQueue.pop();
 87                 char c = nblQueue.poll().charAt(0);
 88                 int res = compute(b,a,c);
 89                 numFromQueue.push(res);
 90             }
 91         }
 92         return (int)numFromQueue.pop();
 93     }
 94     
 95     public static boolean judge(){
 96         String str = nblQueue.peek();
 97         char c = str.charAt(0);
 98         if(c>='0' && c<='9'){
 99             return true;
100         }else{
101             return false;
102         }
103     }
104     
105     //简单加减乘除计算
106         public static int compute(int a, int b, char c){
107             int result = 0;
108             switch(c){
109             case '+': result = a+b;
110                 break;
111             case '-': result = a-b;
112                 break;
113             case '*': result = a*b;
114                 break;
115             case '/': result = a/b;
116                 break;
117             }
118             return result;
119         }
120 
121 }

 

posted @ 2015-09-18 01:47  CHEN0958  阅读(262)  评论(0)  编辑  收藏  举报