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.
1 class Solution { 2 public: 3 int calculate(string s) { 4 stack<int> stack; 5 int res = 0; 6 int len = s.length(); 7 for (int i = 0; i < len; i++) { 8 char c = s[i]; 9 if (isdigit(c)) { 10 int cur = c - ‘0‘; 11 while (i + 1 < len && isdigit(s[i + 1])) { 12 cur = cur * 10 + s[i + 1] - ‘0‘; 13 i++; 14 } 15 stack.push(cur); 16 } else if (c == ‘+‘) { 17 if (stack.size() == 1) 18 stack.push(1); 19 else { 20 res = stack.top(); 21 stack.pop(); 22 res = res * stack.top(); 23 stack.pop(); 24 res = res + stack.top(); 25 stack.pop(); 26 stack.push(res); 27 stack.push(1); 28 } 29 } else if (c == ‘-‘) { 30 if (stack.size() == 1) 31 stack.push(-1); 32 else { 33 res = stack.top(); 34 stack.pop(); 35 res = res * stack.top(); 36 stack.pop(); 37 res = res + stack.top(); 38 stack.pop(); 39 stack.push(res); 40 stack.push(-1); 41 } 42 } else if (c == ‘*‘) { 43 //avoid space 44 while (i + 1 < len && !isdigit(s[i + 1])) { 45 i++; 46 } 47 int cur = s[++i] - ‘0‘; 48 while (i + 1 < len && isdigit(s[i + 1])) { 49 cur = cur * 10 + s[i + 1] - ‘0‘; 50 i++; 51 } 52 res = stack.top() * cur; 53 stack.pop(); 54 stack.push(res); 55 } else if (c == ‘/‘) { 56 while (i + 1 < len && s[i + 1] == ‘ ‘) { 57 i++; 58 } 59 int cur = s[++i] - ‘0‘; 60 while (i + 1 < len && isdigit(s[i + 1])) { 61 cur = cur * 10 + s[i + 1] - ‘0‘; 62 i++; 63 } 64 res = stack.top() / cur; 65 stack.pop(); 66 stack.push(res); 67 } 68 } 69 if (stack.size() == 1) 70 return stack.top(); 71 else { 72 res = stack.top(); 73 stack.pop(); 74 res = res * stack.top(); 75 stack.pop(); 76 res = res + stack.top(); 77 return res; 78 } 79 } 80 };
原文：http://www.cnblogs.com/wxquare/p/5064966.html