Leetcode[224] Basic Calculator

Leetcode[224] Basic Calculator

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ),
the plus +, minus sign - or * or /, non-negative integers and empty spaces .

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23
"1 + (2 - 3 * 4)" = -9;

Stack

复杂度
O(N), O(N)

思路
将字符串先转换成后缀表达式，再将其evaluate出来。
前后缀表达式的转换：
http://scriptasylum.com/tutor...
后缀表达式的evaluation:
http://scriptasylum.com/tutor...

代码

public int basicCalculator(String s) {
    s = s.trim().replaceAll(" +", "");
    Stack<Character> stack = new Stack<>();
    StringBuilder builder = new StringBuilder();
    char[] arr = s.toCharArray();
    for(int i = 0; i < arr.length; i ++) {
        if(arr[i] == '(') {
            stack.push('(');
        }
        else if(arr[i] == ')') {
            while(!stack.isEmpty() && stack.peek() != '(') {
                builder.append(stack.pop());
            }
            // pop out the '('
            stack.pop();
        }
        else if(Character.isDigit(arr[i])) {
            int val = 0;
            for(int j = i; j < arr.length && Character.isDigit(arr[j]); j ++) {
                val *= 10;
                val += arr[j] - '0';
            }
            builder.append(val);
            i = j - 1;
        }
        else {
            while(!stack.isEmpty() && rank(stack.peek()) >= rank(arr[j])) {
                builder.append(stack.pop());
            }
            stack.push(arr[j]);
        }
    }
    while(!stack.isEmpty()) {
        builder.append(stack.pop());
    }
    return evaluate(builder.toString());
}

public int rank(Character ch) {
    if(ch == '+' || ch == '-') return 1;
    else if(ch == '*' || ch == '/') return 2;
    // '('
    return 0;
}

public int evaluate(String s) {
    char[] arr = s.toCharArray();
    Stack<Integer> stack = new Stack<>();
    for(int i = 0; i < arr.length; i ++) {
        if(Character.isDigit(arr[i])) {
            stack.push(arr[i] - '0');
        }
        else {
            int op2 = stack.pop();
            int op1 = stack.pop();
            if(arr[i] == '+') {
                stack.push(op1 + op2);
            }
            else if(arr[i] == '-') {
                stack.push(op1 - op2);
            }
            else if(arr[i] == '*') {
                stack.push(op1 * op2);
            }
            else {
                stack.push(op1 / op2);
            }
        }
    }
    return stack.pop();
}