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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.
用到stack，如果遇到左括号‘(‘将之前的值和1.-1表示运算符号压入stack中，遇到右括号‘)‘，将运算符号和值去除做运算。
1 class Solution { 2 public: 3 int calculate(string s) { 4 if(s.length()==0) return 0; 5 stack<int> mystack; 6 int result=0; 7 int op=1; 8 for(int i=0;i<s.length();i++) 9 { 10 if(s[i]>=‘0‘ && s[i]<=‘9‘) 11 { 12 int number=s[i]-‘0‘; 13 while((i+1)<s.length() && s[i+1]>=‘0‘ && s[i+1]<=‘9‘) 14 { 15 i++; 16 number=number*10+s[i]-‘0‘; 17 } 18 result+=op*number; 19 } 20 if(s[i]==‘ ‘) 21 { 22 continue; 23 } 24 if(s[i]==‘+‘) 25 { 26 op=1; 27 } 28 if(s[i]==‘-‘) 29 { 30 op=-1; 31 } 32 if(s[i]==‘(‘) 33 { 34 mystack.push(result); 35 result=0; 36 mystack.push(op); 37 op=1; 38 } 39 if(s[i]==‘)‘) 40 { 41 op=mystack.top(); 42 mystack.pop(); 43 result=op*result+mystack.top(); 44 mystack.pop(); 45 op=1; 46 } 47 } 48 return result; 49 } 50 };
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原文地址：http://www.cnblogs.com/Sean-le/p/4734608.html
