力扣-150-逆波兰表达式求值

class Solution {
    public int evalRPN(String[] tokens) {
        int len = tokens.length;
        Stack<Integer> stack = new Stack<>();

        for(int i = 0; i < len; i++) {
            Integer x = 0;
            Integer y = 0;
            switch (tokens[i]) {
                case "+":
                    x = stack.pop();
                    x += stack.pop();
                    stack.push(x);
                    break;
                case "-":
                    x = - stack.pop();
                    x += stack.pop();
                    stack.push(x);
                    break;
                case "*":
                    x = stack.pop();
                    x *= stack.pop();
                    stack.push(x);
                    break;
                case "/":
                    x = stack.pop();
                    y = stack.pop();
                    x = y/x;
                    stack.push(x);
                    break;
                default:
                    stack.push(Integer.parseInt(tokens[i]));
                    break;
            }
        }
        return stack.peek();
    }
}

调试了半天，原来是没加break，Java的switch-case语句用的少，挑了个寂寞。

这里给出一个比较好的链接

总之switch case执行时，一定会先进行匹配，匹配成功返回当前case的值，再根据是否有break，判断是否继续输出，或是跳出判断。