224. Basic Calculator I && II && III

Basic Calculator I

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open ( and closing parentheses ), the plus + or minus sign -, 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

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

分析：这题因为不存在乘法和除法，所以，对于里面的减号，我们可以把它当成+（-num）来处理。所以，每次遇到一个数字的时候，我们需要知道这个数字的符号，每当我们把这个数字所有的digit都拿到以后，就可以得到这个数，然后把这个数加到之前的临时结果里。

对于比较特殊的处理是括号，但是这里有一个很巧的思路，我们可以把括号里的表达式call当前的方法来计算。

class Solution {
    public int calculate(String s) {
        int res = 0, num = 0, sign = 1, n = s.length();
        for (int i = 0; i < n; ++i) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9') {
                num = 10 * num + (c - '0');
            } else if (c == '(') {
                int j = i, cnt = 0;
                for (; i < n; ++i) {
                    char letter = s.charAt(i);
                    if (letter == '(') ++cnt;
                    if (letter == ')') --cnt;
                    if (cnt == 0) break;
                }
                num = calculate(s.substring(j + 1, i));
            }
            if (c == '+' || c == '-' || i == n - 1) {
                res += sign * num;
                num = 0;
                sign = (c == '+') ? 1 : -1;
             } 
        }
        return res;
    }
}

 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.

分析：因为没有括号，所以我们可以把加或者减的部分当成一个数，比如 5-2，把它当成（5）+（-2）。同理，对于有乘或者除，或者既有乘又有除的话，也把它当成一个数，比如5-3*2/4=（5）-（3*2/4）。对于乘法和除法，我们总是从左算到右，所以我们可以把* or /之前的部分先存下来，当符号是 * or /的时候，再取出来就可以了。

注意：我们处理的时候，总是处理之前一个符号，而不是当前符号

class Solution {
    public int calculate(String s) {
        int res = 0, num = 0, n = s.length();
        char op = '+';
        Stack<Integer> st = new Stack<>();
        for (int i = 0; i < n; ++i) {
            char ch = s.charAt(i);
            if (Character.isDigit(ch)) {
                num = num * 10 + ch - '0';
            } else if (isOperator(ch)) {
                addToStack(op, num, st);
                op = ch;
                num = 0;
            }
        }
        // handle last case
        addToStack(op, num, st);
        while (!st.empty()) {
            res += st.pop();
        }
        return res;
    }
    
    private boolean isOperator(char ch) {
        if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
            return true;
        }
        return false;
    }
    
    private void addToStack(char op, int num, Stack<Integer> st) {
        if (op == '+') st.push(num);
        if (op == '-') st.push(-num);
        if (op == '*' || op == '/') {
            int tmp = (op == '*') ? st.pop() * num : st.pop() / num;
            st.push(tmp);
        }
    }
}

class Solution {
    public int calculate(String s) {
        int num = 0;
        char operator = '+';
        int last = 0, sum = 0;

        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);

            if (Character.isDigit(c)) {
                num = num * 10 + (c - '0');
            } else if (isOperator(c) || i == s.length() - 1) {
                if (operator == '+') {
                    sum += last;
                    last = num;
                }
                else if (operator == '-') {
                    sum += last;
                    last = -num;
                }
                else if (operator == '*') last *= num;
                else if (operator == '/') last /= num;

                num = 0;
                operator = c;
            }
        }

        return sum += last;
    }

    private boolean isOperator(char c) {
        return c == '+' || c == '-' || c == '*' || c == '/';
    }
}

　　

 Basic Calculator III

Implement a basic calculator to evaluate a simple expression string.

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

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

You may assume that the given expression is always valid. All intermediate results will be in the range of [-2147483648, 2147483647].

Some examples:

"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12

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

分析：只要把括号部分的处理加进来就可以了。

class Solution {
    public int calculate(String s) {
        int res = 0, num = 0, n = s.length();
        char op = '+';
        Stack<Integer> st = new Stack<>();
        for (int i = 0; i < n; ++i) {
            char ch = s.charAt(i);
            if (Character.isDigit(ch)) {
                num = num * 10 + ch - '0';
            } else if (ch == '(') {
                int j = i, cnt = 0;
                for (; i < n; ++i) {
                    char letter = s.charAt(i);
                    if (letter == '(') ++cnt;
                    if (letter == ')') --cnt;
                    if (cnt == 0) break;
                }
                num = calculate(s.substring(j + 1, i));
            } else if (isOperator(ch)) {
                addToStack(op, num, st);
                op = ch;
                num = 0;
            }
        }
        // handle last case
        addToStack(op, num, st);
        while (!st.empty()) {
            res += st.pop();
        }
        return res;
    }

    private static boolean isOperator(char ch) {
        if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {
            return true;
        }
        return false;
    }

    private static void addToStack(char op, int num, Stack<Integer> st) {
        if (op == '+') st.push(num);
        if (op == '-') st.push(-num);
        if (op == '*' || op == '/') {
            int tmp = (op == '*') ? st.pop() * num : st.pop() / num;
            st.push(tmp);
        }
    }
}