leetcode 逆波兰式求解
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
import java.util.Stack; public class Solution { public int evalRPN(String[] tokens) { Stack<Integer> stack = new Stack<Integer>(); int tmp; for(int i = 0; i<tokens.length; i++){ if("+".equals(tokens[i])){ int a = stack.pop(); int b = stack.pop(); stack.add(b+a); } else if("-".equals(tokens[i])){ int a = stack.pop(); int b = stack.pop(); stack.add(b-a); } else if("*".equals(tokens[i])){ int a = stack.pop(); int b = stack.pop(); stack.add(b*a); } else if("/".equals(tokens[i])){ int a = stack.pop(); int b = stack.pop(); stack.add(b/a); } else{ stack.add(Integer.parseInt(tokens[i])); } } return stack.pop(); } }
根据你波兰式求值。看到逆波兰式可以想到栈,扫描表达式,遇到数字则将数字入栈,遇到运算符,时,则从栈顶弹出两个元素,后弹出的元素在运算符的左边,先弹出的元素在元素符的右边,执行运算,将结果入栈。扫描结束后,栈中的元素只剩下一个,即逆波兰式的值
时间: 2024-10-12 08:12:32