leetcode150. 逆波兰表达式求值_leetcode150 逆波兰-优快云博客

示例 1：

输入: ["2", "1", "+", "3", "*"]
输出: 9
解释: ((2 + 1) * 3) = 9
示例 2：

输入: ["4", "13", "5", "/", "+"]
输出: 6
解释: (4 + (13 / 5)) = 6

(a+b)*c-(a+b)/e的后缀表达式为：
(a+b)*c-(a+b)/e
→((a+b)*c)((a+b)/e)-
→((a+b)c*)((a+b)e/)-
→(ab+c*)(ab+e/)-
→ab+c*ab+e/-

思路：

首先声明一个栈stack，然后线性扫描input数组，

如果当前元素是运算符，就把栈的最上面两个元素进行相应运算，把它们俩弹出来，再把运算结果push进栈，

如果当前元素不是运算符，直接push就行了。

注意最后返回的结果要转换成int的形式，

而且除法需要只保留整数位。

python

class Solution(object):
    def evalRPN(self, tokens):
        """
        :type tokens: List[str]
        :rtype: int
        """
        stack = []
        for item in tokens:
            if item in ["+", "-", "*", "/"]:
                if item == "+":
                    temp = int(stack[-2]) + int(stack[-1])
                elif item == "-":
                    temp = int(stack[-2]) - int(stack[-1])
                    # print temp                
                elif item == "*":
                    temp = int(stack[-2]) * int(stack[-1])             
                elif item == "/":
                    temp = int(float(stack[-2])/ float(stack[-1]))                    
                stack.pop()
                stack.pop()
                stack.append(temp)
                
            else:
                stack.append(item)
                
        return int(stack[0])

class Solution {
    public int evalRPN(String[] tokens) {
      Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < tokens.length; i++) {
            switch (tokens[i]) {
                case "+":
                    stack.push(stack.pop() + stack.pop());
                    break;
                case "-":
                    int one = stack.pop();
                    int two = stack.pop();
                    stack.push(two - one);
                    break;
                case "*":
                    stack.push(stack.pop() * stack.pop());
                    break;
                case "/":
                    int t1 = stack.pop();
                    int t2 = stack.pop();
                    stack.push(t2 / t1);
                    break;
                default:
                    stack.push(Integer.parseInt(tokens[i]));
                    break;
            }
        }
        return stack.pop();  
    }
}

逆波兰式https://baike.baidu.com/item/%E9%80%86%E6%B3%A2%E5%85%B0%E5%BC%8F/128437

leetcode150. 逆波兰表达式求值

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