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

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