227.Basic-Calculator-II

227. Basic Calculator II

题目地址

https://leetcode.com/problems/basic-calculator-ii/description/

题目描述

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

Example 1:

Input: "3+2*2"
Output: 7
Example 2:

Input: " 3/2 "
Output: 1
Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:
You may assume that the given expression is always valid.
Do not use the eval built-in library function.

代码

Approach 1:

用栈把每个数字存起来，遇到* /弹出计算再push

每个数字用下个操作符做分割

class Solution {
    public int calculate(String s) {
        if (s == null || s.length() == 0) return 0;
        int len = s.length();
        Stack<Integer> stack = new Stack<Integer>();
        int num = 0;
        char sign = '+';
        for (int i = 0; i < len; i++) {
            if (Character.isDigit(s.charAt(i))) {
                num = num * 10 + s.charAt(i) - '0';
            } 

            if (!Character.isDigit(s.charAt(i)) 
                && s.charAt(i) != ' ' 
                || i == len - 1) { // The last character run twice

                if (sign == '+') {
                    stack.push(num);
                } else if (sign == '-') {
                    stack.push(-num);
                } else if (sign == '*') {
                    int val = stack.pop();
                    stack.push(val * num);
                } else if (sign == '/') {
                    int val = stack.pop();
                    stack.push(val / num);
                }

                sign = s.charAt(i);
                num = 0;
            }
        }

        int ans = 0;
        for (int val : stack) {
            ans += val;
        }

        return ans;
    }
}