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类似这样的后缀表达式: 叫做逆波兰表达式
["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 ["4", "13", "5", "/", "+"] -> (4 + (5 / 13)) -> 4
编译器比较喜欢从栈(stack)里面 pop 两个对象出来计算 然后 继续push 回去, 然后继续这样子计算下去..
算法也很简单
public static void main(String[] args) {
System.out.println(calRPN("4", "13", "5", "/", "+"));
}
public static int calRPN(String... tokens) {
int ret = 0;
String operators = "+-*/";
Stack<String> stack = new Stack<String>();
for (String t : tokens) {
if (!operators.contains(t)) {
stack.push(t);
}else {
int a = Integer.valueOf(stack.pop());
int b = Integer.valueOf(stack.pop());
int v = 0;
if (t.equals("+")) {
v = a + b;
}else if (t.equals("-")) {
v = a - b;
}else if (t.equals("*")) {
v = a * b;
}else if (t.equals("/")) {
v = a / b;
}
stack.push(String.valueOf(v));
}
}
ret = Integer.valueOf(stack.pop());
return ret;
}
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原文地址：http://www.cnblogs.com/jarrah/p/4377492.html
