牛客NC216 逆波兰表达式求值【中等栈 C++/Java/Go/PHP】

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题目

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题目链接：https://www.nowcoder.com/practice/885c1db3e39040cbae5cdf59fb0e9382

核心

栈

参考答案C++

class Solution {public:/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param tokens string字符串vector* @return int整型*/int evalRPN(vector<string>& tokens) {//栈。vector模拟vector<int> stack(tokens.size());int idx = 0;for(int i=0;i< tokens.size();i++){string s = tokens[i];if(s=="+" || s=="-" || s== "*" || s=="/"){int num1 = stack[idx-1];int num2 = stack[idx-2];int cur = 0;if(s=="+"){cur = num2+num1;}if(s=="-"){cur = num2-num1;}if(s=="*"){cur = num2*num1;}if(s=="/"){cur = num2/num1;}stack[idx-2] = cur;idx-=1;}else{stack[idx++] = std::stoi(s);}}return stack[idx-1];}
};

参考答案Java

import java.util.*;public class Solution {/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param tokens string字符串一维数组* @return int整型*/public int evalRPN (String[] tokens) {//栈List<String> strings = Arrays.asList("+", "-", "*", "/");Stack<Integer> stack = new Stack<>();for (String s : tokens) {if (strings.contains(s)) {compute(stack, s);} else {stack.add(Integer.valueOf(s));}}return stack.pop();}public void compute(Stack<Integer> stack, String op) {int num1 = stack.pop();int num2 = stack.pop();int ans = 0;if (op.equals("+"))ans = num2 + num1;if (op.equals("-"))ans = num2 - num1;if (op.equals("*"))ans = num2 * num1;if (op.equals("/"))ans = num2 / num1;stack.add(ans);}
}

参考答案Go

package mainimport "strconv"/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param tokens string字符串一维数组* @return int整型*/
func evalRPN(tokens []string) int {//栈。本答案数组模拟栈stack := make([]int, len(tokens))idx := 0for i := 0; i < len(tokens); i++ {s := tokens[i]if s == "+" || s == "-" || s == "*" || s == "/" {//每次从栈中取出栈顶的2个数，计算结果后存进栈中num1 := stack[idx-1]num2 := stack[idx-2]cur := 0if s == "+" {cur = num2 + num1}if s == "-" {cur = num2 - num1}if s == "*" {cur = num2 * num1}if s == "/" {cur = num2 / num1}idx -= 2stack[idx] = curidx += 1} else {num, _ := strconv.Atoi(s)stack[idx] = numidx++//fmt.Println(stack)}}return stack[idx-1]
}

参考答案PHP

<?php/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可** * @param tokens string字符串一维数组 * @return int整型*/
function evalRPN( $tokens )
{// PHP中数组也是栈。本答案用数组模拟栈$stack =[];$idx = 0;for($i=0;$i<count($tokens);$i++){$s= $tokens[$i];if($s =='+' || $s == '-' || $s =='*' || $s =='/'){//取出栈顶2个数计算结果后存入栈中$num1 = $stack[$idx-1];$num2 = $stack[$idx-2];$cur = 0;if($s =='+') {$cur = $num2+$num1;}if($s =='-') {$cur = $num2-$num1;}if($s =='*') {$cur = $num2*$num1;}if($s =='/') {$cur =intval( $num2/$num1);}$stack[$idx-2] = $cur;$idx-=1;}else{$stack[$idx++] = $s;}}return $stack[$idx-1];
}