Basic Calculator

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

下面是将中序表达式转成后缀表达式的方法，自己写的：

public class Solution {    public boolean isNum(char c){        return c != '(' && c != ')' && c != '+' && c != '-' && c != ' ';    }    public boolean highPriority(char insatck, char next){//the problem involve '+' '-' and '()'if(next == '('){return true;}return false;}public void calcul(Stack<Integer> nums,char operate){int num;if(operate == '+'){num = nums.pop() + nums.pop();nums.push(num);}if(operate == '-'){int num1 = nums.pop();int num2 = nums.pop();num = num2 - num1;nums.push(num);}}    public int calculate(String s) {         int len = s.length();        char[] str = s.toCharArray();        int i = 0;        Stack<Character> stack = new Stack<Character>();        Stack<Integer> nums = new Stack<Integer>();                while(i < len){        while(i<len && str[i] == ' '){        ++i;        }        if(i < len && !isNum(str[i])){        if(stack.isEmpty()){        stack.push(str[i]);                }else{        if(highPriority(stack.peek(),str[i])){        stack.push(str[i]);        }else{//        calculate first then push operate into stack        char operate;        if(str[i] != ')'){        while(!stack.isEmpty() && !highPriority(stack.peek(),str[i]) && stack.peek() != '('){        operate = stack.pop();        calcul(nums,operate);        }        stack.push(str[i]);        }else{        while(!stack.isEmpty() && (operate = stack.pop()) != '('){        calcul(nums,operate);        }        }        }        }        ++i;        }                if(i < len && isNum(str[i])){        int num = 0;        while(i < len && isNum(str[i])){            num = (str[i] - '0') + num * 10;            ++i;            }        nums.push(num);        }        }                while(!stack.isEmpty()){        calcul(nums,stack.pop());        }        return nums.pop();    }}

提交完后看了下discuss，发现其他奇怪解法，有个将字符串逆序的方法，奇淫巧技！！！链接：

https://discuss.leetcode.com/topic/15816/iterative-java-solution-with-stack

又重写了这题，感觉上次代码写的不清楚，这次加了注释

public class Solution {    int get_num(char[] str, int i, int[] len){        int next_num = 0;        while(i < str.length && str[i] <= '9' && str[i] >= '0'){            next_num = next_num*10 + str[i] - '0';            ++len[0];            ++i;        }        return next_num;    }    public int calculate(String s) {       Stack<Character> operate = new Stack<Character>();       Stack<Integer> nums = new Stack<Integer>();       char[] str = s.toCharArray();              for(int i = 0; i < str.length; ++i){       if(str[i] == ' ')continue;                  //只有'('有更高优先级，高优先级或者operate为空时直接压入堆栈       if(str[i] == '('){       operate.push(str[i]);       }              //低优先级或者相等优先级，先取operate栈顶操作符，计算得到的数值压入nums堆栈，并把新操作符压入operate,栈顶是'('例外，此时直接压入操作符       else if(str[i] == '+' || str[i] == '-'){       if((!operate.isEmpty() && operate.peek() == '(')|| operate.isEmpty()){       operate.push(str[i]);       }else{       char ope = operate.pop();       int num1 = nums.pop();       int num2 = nums.pop();       if(ope == '-')nums.push(num2 - num1);       if(ope == '+')nums.push(num2 + num1);       operate.push(str[i]);       }       }              //如果新操作符')'则需先弹出operate栈顶的计算操作符计算后再把'('弹出，如果operate栈顶操作符就是'('，则不需要计算这一步骤,直接对operate栈pop       else if(str[i] == ')'){       if(operate.peek() == '('){       operate.pop();       }else{       char ope = operate.pop();       int num1 = nums.pop();       int num2 = nums.pop();       if(ope == '-')nums.push(num2 - num1);       if(ope == '+')nums.push(num2 + num1);       operate.pop();       }       }       else{       int[] len = new int[]{0};       nums.push(get_num(str, i, len));       i = i + len[0] - 1;       }       }       if(!operate.isEmpty()){       char ope = operate.pop();   int num1 = nums.pop();   int num2 = nums.pop();   if(ope == '-')nums.push(num2 - num1);   if(ope == '+')nums.push(num2 + num1);       }       return nums.pop();    }}