227. Basic Calculator II

题目：

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.

链接： http://leetcode.com/problems/basic-calculator-ii/

7/23/2017

6%...

很僵化的只会先infix to postfix，再来计算

注意：

1. string去掉white space: s.replaceAll("\\s+", "")

2. 各种类型的转换, String <--> Integer, String <--> Character...

3. Stack, Queue的接口函数，已经Queue的2套函数都要熟练

4. infix到postfix如何转换？运算数直接输出，运算符要跟stack最上面判断，如果stack top的优先级高或者有相同优先级，要把当前top的运算符输出。无论当前是否有更高的优先级，当前运算符必须要加到stack里面去，因为第二个运算数还没有处理到。最后要把最后一个运算数和所有stack里的运算符输出。

5. 运算postfix很简单，遇到运算数放到stack里，遇到运算符取出stack最上面2个运算数，将结果压回stack。最后stack中只剩下一个数，即为结果。

public class Solution {
    public int calculate(String s) {
        // remove white spaces from input string
        s = s.replaceAll("\\s+","");
        Stack<String> operators = new Stack<>();
        Queue<String> postfix = new LinkedList<>();
        
        StringBuilder operand = new StringBuilder();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            // build operands from one or more characters
            if (c >= '0' && c <= '9') {
                operand.append(c);
            } else {
                // put last operand to output
                postfix.offer(operand.toString());
                operand = new StringBuilder();
                // add operator to operators stack
                if (!operators.isEmpty()) {
                    String prevOperator;
                    while (!operators.isEmpty()) {
                        prevOperator = operators.peek();
                        // if operator on top of stack has same or higher precedence, move top to output
                        if (prevOperator.equals("*") || prevOperator.equals("/") || c == '+' || c == '-') {
                            postfix.offer(operators.pop());
                        } else {
                            break;
                        }
                    }
                }
                // add current operator to operators stack, because the second operand is not visited yet
                operators.push(Character.toString(c));
            }
        }
        // last operand not in postfix yet
        postfix.offer(operand.toString());
        // put all operators stack to postfix
        while (!operators.isEmpty()) {
            postfix.offer(operators.pop());
        }
        // calculate postfix
        Stack<Integer> operandStack = new Stack<>();
        while (!postfix.isEmpty()) {
            String o = postfix.poll();
            if (o.equals("+") || o.equals("-") || o.equals("*") || o.equals("/")) {
                Integer operand2 = operandStack.pop();
                Integer operand1 = operandStack.pop();
                Integer result;
                if (o.equals("+")) {
                    result = operand1 + operand2;
                } else if (o.equals("-")) {
                    result = operand1 - operand2;
                } else if (o.equals("*")) {
                    result = operand1 * operand2;
                } else {
                    result = operand1 / operand2;
                }
                operandStack.push(result);
            } else {
                operandStack.push(Integer.valueOf(o));
            }
        }
        return operandStack.peek();
    }
}

别人的答案

高优先级直接运算，低优先级存运算数

https://discuss.leetcode.com/topic/16935/share-my-java-solution

看不太懂，看解释会好一点

https://discuss.leetcode.com/topic/16807/17-lines-c-easy-20-ms

发现大家都是按照第一种做法来的，直接将中间值保存起来，这个都没有用到stack

https://discuss.leetcode.com/topic/17435/java-straight-forward-iteration-solution-with-comments-no-stack-o-n-o-1

将operand, operator放到2个不同的stack里

https://discuss.leetcode.com/topic/17213/my-28ms-c-code-with-two-stacks-one-for-op-one-for-oprand-extension-to-cover-also-given

有详细解释

https://discuss.leetcode.com/topic/41118/simple-c-solution-beats-85-submissions-with-detailed-explanations

更多讨论

https://discuss.leetcode.com/category/235/basic-calculator-ii