标签：opera expr other else let span ken operator param
Evaluate the value of an arithmetic expression in Reverse Polish Notation.
Valid operators are
+,-,*,/. Each operand may be an integer or another expression.
Some examples:
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6
用栈来做，如果是数字则push进去，如果遇到运算符则将栈顶的两个数字push出，运算完成后再push进栈。
我的JavaScript代码：
var evalRPN = function(tokens) { var arr = []; var num; for(let i = 0; i < tokens.length; i++){ if( !isNaN(tokens[i]) ){ arr.push(tokens[i]); }else{ let a = arr.pop(),b = arr.pop(); a = parseInt(a); b = parseInt(b); if( tokens[i] === ‘+‘ ){ num = b+a; }else if( tokens[i] === ‘-‘ ){ num = b-a; }else if( tokens[i] === ‘*‘ ){ num = b*a; }else if( tokens[i] === ‘/‘ ){ if( (a*b)<0 ) num = Math.ceil( b/a ); else num = Math.floor( b/a ); } arr.push( num ); } } return parseInt( arr[0] ); };
参考了https://discuss.leetcode.com/topic/18179/accepted-clean-java-solution之后，使用swith-case，更加干净的代码：
/** * @param {string[]} tokens * @return {number} */ var evalRPN = function(tokens) { var stack = []; for (var i = 0; i < tokens.length; i++) { switch (tokens[i]) { case "+": stack.push(stack.pop() + stack.pop()); break; case "-": stack.push(-stack.pop() + stack.pop()); break; case "*": stack.push(stack.pop() * stack.pop()); break; case "/": var n1 = stack.pop(), n2 = stack.pop(); if( n1*n2 > 0) stack.push( Math.floor(n2 / n1) ); else stack.push( Math.ceil( n2 / n1) ); break; default: stack.push( parseInt(tokens[i]) ); } } return stack.pop(); };
1. 判断字符串为数字的方法：isNaN( str );
2. 除法与减法操作时，数字的先后位置需要注意;
3. JavaScript中，整除的操作需要分为两类：
150. Evaluate Reverse Polish Notation
标签：opera expr other else let span ken operator param
原文地址：http://www.cnblogs.com/bubbleStar/p/6119922.html
