【LeetCode】Basic Calculator && Basic Calculator II

1、Basic Calculator

Total Accepted: 3726 Total Submissions: 24053 My Submissions Question Solution

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.

2、Basic Calculator II

Total Accepted: 1046 Total Submissions: 5287 My Submissions Question Solution

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.

【解题思路】

1带括号，不带乘除。2不带括号，带乘除

下一个升级版有可能是带括号又带乘除。

1、每次计算都需要判断当前符号是不是括号，如果是括号就先算括号内的值。在discuss中看到一个很漂亮的代码。参考Iterative Java solution with stack

2、遇乘除计算。然后按照顺序计算一次加减的值。

1、Java AC

public class Solution {
    public int calculate(String s) {
        Stack<Integer> stack = new Stack<Integer>();
        int len = s.length();
        int result = 0;
        int sign = 1;
        int num = 0;
        for (int i = 0; i < len; i++) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9') {
                num = 10 * num + c - '0';
            } else if (c == '+') {
                result += sign * num;
                sign = 1;
                num = 0;
            } else if (c == '-') {
                result += sign * num;
                sign = -1;
                num = 0;
            } else if (c == '(') {
                stack.push(result);
                stack.push(sign);
                result = 0;
                sign = 1;
            } else if (c == ')') {
                result += sign * num;
                num = 0;
                result *= stack.pop();
                result += stack.pop();
            }
        }
        if (num != 0) {
            result += sign * num;
        }
        return result;
    }
}

2、Java AC

public class Solution {
    public int calculate(String s) {
        if (s == null || "".equals(s.trim())) {
            return 0;
        }
        s = s.replaceAll("\\s", "");
        List<Long> list1 = new ArrayList<Long>();
        List<Character> list2 = new ArrayList<Character>();
        int len = s.length();
        int i = 0;
        while (i < len) {
            char c = s.charAt(i);
            if (c >= '0' && c <= '9') {
                long array[] = getNum(s, i);
                list1.add(array[0]);
                i = (int) array[1];
            } else if (c == '+' || c == '-') {
                list2.add(c);
            } else if (c == '*' || c == '/') {
                int size = list1.size();
                long num1 = list1.get(size - 1);
                long array[] = getNum(s, i + 1);
                long num2 = array[0];
                i = (int) array[1];
                long num = c == '*' ? num1 * num2 : num1 / num2;
                list1.set(size - 1, num);
            }
            i++;
        }
        int size1 = list1.size();
        int size2 = list2.size();
        long preNum = list1.get(0);
        for (i = 0; i < size2; i++) {
            if (list2.get(i) == '+') {
                preNum += list1.get(i + 1);
            } else {
                preNum -= list1.get(i + 1);
            }
        }
        return (int) preNum;
    }

    private long[] getNum(String s, int curPos) {
        StringBuffer sb = new StringBuffer();
        int len = s.length();
        while (curPos < len && s.charAt(curPos) >= '0'
                && s.charAt(curPos) <= '9') {
            sb.append(s.charAt(curPos));
            curPos++;
        }
        String newStr = sb.toString();
        if (newStr.length() > 0) {
            return new long[]{Long.parseLong(newStr), curPos - 1};
        }
        return new long[]{0, curPos - 1};
    }
}