LeetCode 227 Basic Calculator II

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.

You may assume that the given expression is always valid.

Some examples:

"3+2*2" = 7" 3/2 " = 1" 3+5 / 2 " = 5

Note: Do not use the eval built-in library function.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

---

分析

分两次遍历，第一次遍历时，遇到乘除符号就计算；第二次遍历，计算加减符号。

参考：http://blog.csdn.net/xudli/article/details/46644317

---

代码

    public static int calculate(String s) {        if (s == null || s.length() == 0) {            return 0;        }        Stack<Integer> stack = new Stack<Integer>();        for (int i = 0; i < s.length(); i++) {            char c = s.charAt(i);            if (Character.isDigit(c)) {                int val = c - '0';                // 将数字取出                while (i + 1 < s.length() && Character.isDigit(s.charAt(i + 1))) {                    val = val * 10 + s.charAt(++i) - '0';                }                // 栈顶是 * / 运算符，计算                if (!stack.isEmpty()                        && (stack.peek() == 2 || stack.peek() == 3)) {                    int sign = stack.pop();                    int op = stack.pop();                    int rt = 0;                    if (sign == 2) {                        rt = op * val;                    } else {                        rt = op / val;                    }                    stack.push(rt);                } else {                    stack.push(val);                }            } else if (c == ' ') {                continue;            } else {                switch (c) {                case '+':                    stack.push(0);                    break;                case '-':                    stack.push(1);                    break;                case '*':                    stack.push(2);                    break;                case '/':                    stack.push(3);                    break;                }            }        }        if (stack.isEmpty()) {            return 0;        }        // 因为要从左向右计算，所以要reverse        Collections.reverse(stack);        // 计算+-        int rt = stack.pop();        while (!stack.isEmpty()) {            int sign = stack.pop();            int op = stack.pop();            if (sign == 0) {                rt += op;            } else {                rt -= op;            }        }        return rt;    }