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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.
一个很详细的解释：http://www.geeksforgeeks.org/expression-evaluation/ 比这道题还难点
思路就是两个stack，一个存数字一个存符号。如果遇到数字直接存到数字stack；如果遇到符号，有几种情况：
1.当前符号比上一个符号优先级高，比如* 高于+，那么直接进栈
2.当前符号低于上一个，那么就要把所有已经在stack里面优先于当前符号的全算完，再推进当前符号
3.当前符号是“（”，直接push
4.当前符号是“）”，就要把所有“（”以前的符号全部算完
这道题只有“+”号与“-”号，不用判断符号的优先级。
1 class Solution { 2 public: 3 int calculate(string s) { 4 stack<int> stk_val; 5 stack<char> stk_op; 6 int res = 0, tmp; 7 for (int i = 0; i <= s.length(); ++i) { 8 //操作数 9 if (i < s.length() && isdigit(s[i])) { 10 res = 0; 11 while (i < s.length() && isdigit(s[i])) { 12 res *= 10; 13 res += s[i++] - ‘0‘; 14 } 15 stk_val.push(res); 16 } 17 //运算符 18 if (i == s.length() || s[i] == ‘+‘ || s[i] == ‘-‘ || s[i] == ‘)‘) { 19 while (!stk_op.empty() && stk_op.top() != ‘(‘) { 20 tmp = stk_val.top(); 21 stk_val.pop(); 22 if (stk_op.top() == ‘+‘) stk_val.top() += tmp; 23 else if (stk_op.top() == ‘-‘) stk_val.top() -= tmp; 24 stk_op.pop(); 25 } 26 if (i == s.length()) break; 27 else if (s[i] == ‘)‘) stk_op.pop(); 28 else stk_op.push(s[i]); 29 } else if (s[i] == ‘(‘) { 30 stk_op.push(s[i]); 31 } 32 } 33 return stk_val.top(); 34 } 35 };
LintCode上有一道相同的题，但是运算符不仅包含"+"、"-"，还包含了 "*" 和 "/"，不过不用自己解析操作数了。
Given an expression string array, return the final result of this expression
For the expression 2*6-(23+7)/(1+2), input is
[
"2", "*", "6", "-", "(",
"23", "+", "7", ")", "/",
(", "1", "+", "2", ")"
],
return 2
The expression contains only integer, +, -, *, /, (, ).
1 class Solution { 2 public: 3 /** 4 * @param expression: a vector of strings; 5 * @return: an integer 6 */ 7 int calulate(int a, int b, const string &op) { 8 if (op == "+") return a + b; 9 else if (op == "-") return a - b; 10 else if (op == "*") return a * b; 11 else return a / b; 12 } 13 bool isOK(const string &op1, const string &op2) { 14 if (op1 == "*" || op1 == "/" || op2 == ")") return true; 15 else return op2 == "+" || op2 == "-"; 16 } 17 int evaluateExpression(vector<string> &expression) { 18 // write your code here 19 if (expression.empty()) return 0; 20 stack<int> stk_val; 21 stack<string> stk_op; 22 for (int i = 0; i <= expression.size(); ++i) { 23 if (i < expression.size() && isdigit(expression[i][0])) { 24 stk_val.push(atoi(expression[i].c_str())); 25 } else if (i == expression.size() || expression[i] != "(") { 26 while (!stk_op.empty() && stk_op.top() != "(" 27 && (i == expression.size() || isOK(stk_op.top(), expression[i]))) { 28 int tmp = stk_val.top(); 29 stk_val.pop(); 30 stk_val.top() = calulate(stk_val.top(), tmp, stk_op.top()); 31 stk_op.pop(); 32 } 33 if (i == expression.size()) break; 34 else if (expression[i] == ")") stk_op.pop(); 35 else stk_op.push(expression[i]); 36 } else { 37 stk_op.push(expression[i]); 38 } 39 } 40 return stk_val.top(); 41 } 42 };
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原文地址：http://www.cnblogs.com/easonliu/p/4563780.html
