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Basic Calculator II Leetcode

2024-09-04 13:19:39   220人阅读

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.

这道题写的不好，写的太麻烦了。。。

public class Solution {
    public int calculate(String s) {
        if (s == null) {
            return 0;
        }
        int result = 0;
        List<String> tmp = new ArrayList<>();
        int num = 0;
        for (int i = 0; i < s.length(); i++) {
            if (Character.isDigit(s.charAt(i))) {
                num = num * 10 + (s.charAt(i) - ‘0‘);
            } else {
                if (s.charAt(i) != ‘ ‘) {
                    tmp.add(String.valueOf(num));
                    num = 0;
                    tmp.add(s.charAt(i) + "");
                }
            }
        }
        tmp.add(String.valueOf(num));
        for (int i = 0; i < tmp.size(); i++) {
            int t = 0;
            if (tmp.get(i).equals("*") || tmp.get(i).equals("/")) {
                int a = Integer.valueOf(tmp.get(i - 1));
                int b = Integer.valueOf(tmp.get(i + 1));
                if (tmp.get(i).equals("*")) {
                    t = a * b;
                } else if (tmp.get(i).equals("/")) {
                    t = a / b;
                }
                tmp.remove(i - 1);
                tmp.remove(i - 1);
                tmp.remove(i - 1);
                tmp.add(i - 1, String.valueOf(t));
                i = i - 2;
            }
        }
        result += Integer.valueOf(tmp.get(0));
        for (int i = 1; i < tmp.size(); i++) {
            switch (tmp.get(i)) {
                case "+":
                    result += Integer.valueOf(tmp.get(i + 1));
                    i++;
                    break;
                case "-" :
                    result -= Integer.valueOf(tmp.get(i + 1));
                    i++;
                    break;
            }
        }
        return result;
    }
}

这效率也是没谁了。。。

技术分享

还是用好的写法再写一遍吧。。。。哭瞎了。。。

看了top solution发现思路还是基本一致的嘛。。。恩。。。。就是我没用stack。。。而且减号可以转换成负数存进去。

思想就是存储的时候就把乘法除法处理好，然后再集体算一遍。注意处理一下遍历到最后的情况。

public class Solution {
    public int calculate(String s) {
        if (s == null) {
            return 0;
        }
        s = s + "+";
        Stack<Integer> stack = new Stack<>();
        char sign = ‘+‘;
        int num = 0;
        for (int i = 0; i < s.length(); i++) {
            if (Character.isDigit(s.charAt(i))) {
                num = num * 10 + (s.charAt(i) - ‘0‘);
            } else {
                if (s.charAt(i) != ‘ ‘) {
                    if (sign == ‘+‘) {
                        stack.push(num);
                    }
                    if (sign == ‘-‘) {
                        stack.push(-num);
                    }
                    if (sign == ‘*‘) {
                        stack.push(num * stack.pop());
                    }
                    if (sign == ‘/‘) {
                        stack.push(stack.pop() / num);
                    }
                    sign = s.charAt(i);
                    num = 0;
                }
            }
        }
        int result = 0;
        while (!stack.isEmpty()) {
            result += stack.pop();
        }
        return result;
    }
}

这个版本在最后加了一个加号来计算最后一个值，还有top solution的版本

public class Solution {
    public int calculate(String s) {
        if (s == null) {
            return 0;
        }
        Stack<Integer> stack = new Stack<>();
        char sign = ‘+‘;
        int num = 0;
        for (int i = 0; i < s.length(); i++) {
            if (Character.isDigit(s.charAt(i))) {
                num = num * 10 + (s.charAt(i) - ‘0‘);
            }
            if ((!Character.isDigit(s.charAt(i)) && s.charAt(i) != ‘ ‘) || i == s.length() - 1) {
                if (sign == ‘+‘) {
                    stack.push(num);
                }
                if (sign == ‘-‘) {
                    stack.push(-num);
                }
                if (sign == ‘*‘) {
                    stack.push(num * stack.pop());
                }
                if (sign == ‘/‘) {
                    stack.push(stack.pop() / num);
                }
                sign = s.charAt(i);
                num = 0;
            }
        }
        int result = 0;
        while (!stack.isEmpty()) {
            result += stack.pop();
        }
        return result;
    }
}

这个稍微慢一点，大概是判断次数比较多吧。

 

Basic Calculator II Leetcode


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