leetcode Basic Calculator II

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.

没有了括号，只需要考虑乘除的高优先级问题，discuss中有一个比较清楚的方法：建立一个栈，当运算符为乘除时，将栈顶元素乘除当前数，当运算符为加减时，直接将该数压入栈顶，最后将栈内的数累加即为所求：

public int calculate(String s) {
    int i=0;
    Stack<Integer> stack=new Stack<>();
    char sign='+';
    while(i<s.length()){
        while(i<s.length()&&s.charAt(i)==' '){
            i++;
        }
        if(i==s.length()) break;
        if(s.charAt(i)=='+'||s.charAt(i)=='-'||s.charAt(i)=='*'||s.charAt(i)=='/') {
            sign=s.charAt(i);
            i++;
        }
        else{
            int k=0;
            while(i<s.length()&&s.charAt(i)>='0'&&s.charAt(i)<='9'){
                k=k*10+(s.charAt(i)-'0');
                i++;
            }
            if(sign=='+') stack.push(k);
            else if(sign=='-') stack.push(-k);
            else if(sign=='*') stack.push(stack.pop()*k);
            else if(sign=='/') stack.push(stack.pop()/k);
        }
    }
    int res=0;
    for(int num:stack){
        res+=num;
    }
    return res;
}