问题描述
输入(在JavaScript)为3-2 +(8-3)
我想翻译这个前pression到逆波兰式。然而,根据算法,我可以得到3 2 8 3 - + - ,它不评估的结果12 .....左右方法的任何工作呢?我知道括号是不必要的位置,但是,哦...我有我下面的功能:
函数ShuntingYard(STR){
海峡= str.replace(/ \)\(/克,)*();
VAR ARR = str.split();
变种sYqueue = [];
变种sYstack = [];
而(arr.length大于0){
VAR令牌= arr.shift();
如果(/ \ D + /。测试(标记)){
//如果它是一个数字,推到队列
sYqueue.push(标记);
如果} //结束
否则,如果(/ [+] | [ - ] | [*] | [\ /] /测试(标记)){
//如果它是一个运营商
如果(sYstack.length == 0){
//如果一个空的运营商栈
sYstack.push(标记);
}
其他{
而((/[*]|[\/]/.test(sYstack[sYstack.length-1]))及&安培;
(/ [+] | [ - ] /测试(标记))){
//如果TOS有运营商,拥有较高的precedence
//那么就需要弹出堆栈
//并添加到队列
执行console.log(sYstack);
sYqueue.push(sYstack.pop());
}
sYstack.push(标记);
}
}
否则,如果(/ [(] /。测试(标记)){
//如果它的左括号
sYstack.push(标记);
}
否则如果(/ [)] /。试验(令牌)){
//如果是右括号
而(!(/ [(] /。试验(sYstack [sYstack.length-1]))){
//虽然没有在堆栈的顶部没有左括号
//那么就需要弹出的运营商到队列中
sYqueue.push(sYstack.pop());
} //结束时
如果(sYstack.length == 0)
{//不平衡的括号!
的console.log(错误,不平衡的括号);
}
其他
{
sYstack.pop(); //关闭弹出左括号
}
}
其他{
//其他情况
}
} //结束时
//现在而栈不为空,弹出每一个运营商要排队
而(sYstack.length大于0){
sYqueue.push(sYstack.pop());
}
返回sYqueue;
} //函数结束ShuntingYard
很久很久以前,在一个依据很远很远我在JavaScript写的Dijkstra的调车场算法的实现:
函数分析器(表){
this.table =表;
}
Parser.prototype.parse =功能(输入){
VAR长度= input.length,
表= this.table,
输出= [],
堆栈= [],
索引= 0;
而(指数<长度){
VAR标记=输入[指数++];
开关(标记){
外壳 (:
stack.unshift(标记);
打破;
外壳 ):
而(stack.length){
VAR令牌= stack.shift();
如果(标记===()打破;
其他output.push(标记);
}
如果(令牌!==()
抛出新的错误(不匹配的括号内。);
打破;
默认:
如果(table.hasOwnProperty(令牌)){
而(stack.length){
VAR标点符号=堆叠[0];
如果(标点符号===()打破;
VAR运算符=表[令牌]
precedence =运算符。precedence,
。先行=表[标点符号] precedence;
如果(precedence>先行||
precedence ===先行和放大器;&安培;
operator.associativity ===右)打破;
别的output.push(stack.shift());
}
stack.unshift(标记);
}其他output.push(标记);
}
}
而(stack.length){
VAR令牌= stack.shift();
如果(令牌==()output.push(标记)!;
否则抛出新的错误(不匹配的括号内。);
}
返回输出;
};
下面是你将如何使用它:
VAR分析器=新的解析器({ *:{precedence:2,关联性:左}, /:{precedence:2,关联性:左}, +:{precedence:1,关联性:左}, - :{precedence:1,关联性:左}});无功输出= parser.parse(3 - 2 +(8 - 3)分裂(。();))加入。警报(JSON.stringify(输出)); //3月二号至8月3号 - +
<脚本>功能分析器(一){this.table =一}分析器.prototype.parse =函数(){VAR B =则为a.length,表= this.table,输出= [],堆栈= [],指数= 0;而(指数< B){VAR C = A [索引++ ];开关(三){案(:stack.unshift(C);打破;案件):而(stack.length){VAR C = stack.shift();如果(C ===( )打破;否则output.push(C)},如果(C = =!()抛出新的错误;打破;默认情况下(不匹配的括号。):如果(table.hasOwnProperty(C)){而(堆栈.length){VAR D =栈[0];如果(D ===()打破; VAR e=table[c],$p$pcedence=e.$p$pcedence,antecedence=table[d].$p$pcedence;if($p$pcedence>antecedence||$p$pcedence===antecedence&&e.associativity==="right")break;else output.push(stack.shift())} stack.unshift(C)}其他output.push(C)}}而(stack.length){VAR C = stack.shift(!);如果(C == ()output.push(C);否则抛出新的错误(不匹配的括号);}返回输出};< / SCRIPT>
这,顺便说一句不(而且永远不会)计算为 12
,但这:
VAR分析器=新的解析器({ *:{precedence:2,关联性:左}, /:{precedence:2,关联性:左}, +:{precedence:1,关联性:左}, - :{precedence:1,关联性:左}});无功输出= parser.parse(3 * 3 - 2 + 8 - 3.split(();))加入。警报(JSON.stringify(输出)); //3 3 * 2 - 8 + 3 -
<脚本>功能分析器(一){this.table =一}分析器.prototype.parse =函数(){VAR B =则为a.length,表= this.table,输出= [],堆栈= [],指数= 0;而(指数< B){VAR C = A [索引++ ];开关(三){案(:stack.unshift(C);打破;案件):而(stack.length){VAR C = stack.shift();如果(C ===( )打破;否则output.push(C)},如果(C = =!()抛出新的错误;打破;默认情况下(不匹配的括号。):如果(table.hasOwnProperty(C)){而(堆栈.length){VAR D =栈[0];如果(D ===()打破; VAR e=table[c],$p$pcedence=e.$p$pcedence,antecedence=table[d].$p$pcedence;if($p$pcedence>antecedence||$p$pcedence===antecedence&&e.associativity==="right")break;else output.push(stack.shift())} stack.unshift(C)}其他output.push(C)}}而(stack.length){VAR C = stack.shift(!);如果(C == ()output.push(C);否则抛出新的错误(不匹配的括号);}返回输出};< / SCRIPT>
有你有它:在JavaScript广义实现Dijkstra的调车场算法
Input (in javascript) is "3-2+(8-3)"
I want to translate this expression to Reverse Polish Notation.However, according to the algorithm, I can get "3 2 8 3 - + -", which doesn't evaluate to the result 12..... Any work around method for this? I know the parentheses are unnecessary here, but, oh well...I have my function below:
function ShuntingYard(str){
str=str.replace(/\)\(/g, ")*(");
var arr=str.split("");
var sYqueue=[];
var sYstack=[];
while (arr.length>0){
var token=arr.shift();
if (/\d+/.test(token)){
// if it's a number, push to the queue
sYqueue.push(token);
} // end if
else if (/[+]|[-]|[*]|[\/]/.test(token)){
// if it's an operator
if (sYstack.length==0){
// if an empty operator stack
sYstack.push(token);
}
else{
while ((/[*]|[\/]/.test(sYstack[sYstack.length-1])) &&
(/[+]|[-]/.test(token))){
// if the TOS has operator with higher precedence
// then need to pop off the stack
// and add to queue
console.log(sYstack);
sYqueue.push(sYstack.pop());
}
sYstack.push(token);
}
}
else if (/[(]/.test(token)){
// if it's left parenthesis
sYstack.push(token);
}
else if (/[)]/.test(token)){
// if it's right parenthesis
while (!(/[(]/.test(sYstack[sYstack.length-1]))){
// while there's no left parenthesis on top of the stack
// then need to pop the operators onto the queue
sYqueue.push(sYstack.pop());
} // end while
if (sYstack.length==0)
{ // unbalanced parenthesis!!
console.log("error, unbalanced parenthesis");
}
else
{
sYstack.pop(); // pop off the left parenthesis
}
}
else{
// other cases
}
} // end while
// now while the stack is not empty, pop every operators to queue
while (sYstack.length>0){
sYqueue.push(sYstack.pop());
}
return sYqueue;
} // end function ShuntingYard
A long time ago in a gist far far away I wrote an implementation of Dijkstra's shunting yard algorithm in JavaScript:
function Parser(table) {
this.table = table;
}
Parser.prototype.parse = function (input) {
var length = input.length,
table = this.table,
output = [],
stack = [],
index = 0;
while (index < length) {
var token = input[index++];
switch (token) {
case "(":
stack.unshift(token);
break;
case ")":
while (stack.length) {
var token = stack.shift();
if (token === "(") break;
else output.push(token);
}
if (token !== "(")
throw new Error("Mismatched parentheses.");
break;
default:
if (table.hasOwnProperty(token)) {
while (stack.length) {
var punctuator = stack[0];
if (punctuator === "(") break;
var operator = table[token],
precedence = operator.precedence,
antecedence = table[punctuator].precedence;
if (precedence > antecedence ||
precedence === antecedence &&
operator.associativity === "right") break;
else output.push(stack.shift());
}
stack.unshift(token);
} else output.push(token);
}
}
while (stack.length) {
var token = stack.shift();
if (token !== "(") output.push(token);
else throw new Error("Mismatched parentheses.");
}
return output;
};
Here is how you would use it:
var parser = new Parser({
"*": { precedence: 2, associativity: "left" },
"/": { precedence: 2, associativity: "left" },
"+": { precedence: 1, associativity: "left" },
"-": { precedence: 1, associativity: "left" }
});
var output = parser.parse("3 - 2 + ( 8 - 3 )".split(" ")).join(" ");
alert(JSON.stringify(output)); // "3 2 - 8 3 - +"
<script>function Parser(a){this.table=a}Parser.prototype.parse=function(a){var b=a.length,table=this.table,output=[],stack=[],index=0;while(index<b){var c=a[index++];switch(c){case"(":stack.unshift(c);break;case")":while(stack.length){var c=stack.shift();if(c==="(")break;else output.push(c)}if(c!=="(")throw new Error("Mismatched parentheses.");break;default:if(table.hasOwnProperty(c)){while(stack.length){var d=stack[0];if(d==="(")break;var e=table[c],precedence=e.precedence,antecedence=table[d].precedence;if(precedence>antecedence||precedence===antecedence&&e.associativity==="right")break;else output.push(stack.shift())}stack.unshift(c)}else output.push(c)}}while(stack.length){var c=stack.shift();if(c!=="(")output.push(c);else throw new Error("Mismatched parentheses.");}return output};</script>
This, incidentally does not (and never will) evaluate to 12
, but this does:
var parser = new Parser({
"*": { precedence: 2, associativity: "left" },
"/": { precedence: 2, associativity: "left" },
"+": { precedence: 1, associativity: "left" },
"-": { precedence: 1, associativity: "left" }
});
var output = parser.parse("3 * 3 - 2 + 8 - 3".split(" ")).join(" ");
alert(JSON.stringify(output)); // "3 3 * 2 - 8 + 3 -"
<script>function Parser(a){this.table=a}Parser.prototype.parse=function(a){var b=a.length,table=this.table,output=[],stack=[],index=0;while(index<b){var c=a[index++];switch(c){case"(":stack.unshift(c);break;case")":while(stack.length){var c=stack.shift();if(c==="(")break;else output.push(c)}if(c!=="(")throw new Error("Mismatched parentheses.");break;default:if(table.hasOwnProperty(c)){while(stack.length){var d=stack[0];if(d==="(")break;var e=table[c],precedence=e.precedence,antecedence=table[d].precedence;if(precedence>antecedence||precedence===antecedence&&e.associativity==="right")break;else output.push(stack.shift())}stack.unshift(c)}else output.push(c)}}while(stack.length){var c=stack.shift();if(c!=="(")output.push(c);else throw new Error("Mismatched parentheses.");}return output};</script>
There you have it: a generalized implementation of Dijkstra's shunting yard algorithm in JavaScript.
这篇关于调车场算法不必要的括号的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!