227. Basic Calculator II

Implement a basic calculator to evaluate a simple expression string.

The expression string contains onlynon-negativeintegers,+,-,*,/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 evalbuilt-in library function.

Analysis

link:

http://www.cnblogs.com/grandyang/p/4601208.html

https://leetcode.com/problems/basic-calculator-ii/discuss/63003/Share-my-java-solution

由于存在运算优先级，我们采取的措施是使用一个栈保存数字，如果该数字之前的符号是加或减，那么把当前数字压入栈中，注意如果是减号，则加入当前数字的相反数，因为减法相当于加上一个相反数。如果之前的符号是乘或除，那么从栈顶取出一个数字和当前数字进行乘或除的运算，再把结果压入栈中，那么完成一遍遍历后，所有的乘或除都运算完了，再把栈中所有的数字都加起来就是最终结果了

Solution

public int calculate(String s) {
    int len;
    if (s == null || (len = s.length()) == 0)
        return 0;
    Deque<Integer> stack = new LinkedList<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) {
            if (sign == '-') {
                stack.push(-num);
            }
            if (sign == '+') {
                stack.push(num);
            }
            if (sign == '*') {
                stack.push(stack.pop() * num);
            }
            if (sign == '/') {
                stack.push(stack.pop() / num);
            }
            sign = s.charAt(i);
            num = 0;
        }
    }

    int re = 0;
    for (int i : stack) {
        re += i;
    }
    return re;
}