15
06/2015
[LeetCode] Basic Calculator
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
Note: Do not use the eval built-in library function.
解题思路：
简易计算器。现将中缀表达式转化成后缀表达式，然后利用栈来计算。有几个注意的地方。1、后缀表达式的转化问题。2、对于减号而言，注意哪个数是减数，哪个是被减数。这个程序很容易修改成包含加减乘除的计算器问题。不同的就是后缀表达式的转化不一样。
class Solution {
public:
int calculate(string s) {
string postS = getPostString(s); //获得后缀表达式
cout << postS;
stack<int> nums;
int len = postS.length();
int num1, num2;
for (int i = 0; i<len; i++){
switch (postS[i])
{
case '+':
num1 = nums.top();
nums.pop();
num2 = nums.top();
nums.pop();
num1 = num1 + num2;
nums.push(num1);
break;
case '-':
num1 = nums.top();
nums.pop();
num2 = nums.top();
nums.pop();
num1 = num2 - num1; //注意这里是num2减去num1
nums.push(num1);
break;
case '#':
break;
default:
int num = 0;
while (i < len && postS[i] >= '0' && postS[i] <= '9'){
num = num * 10 + (postS[i] - '0');
i++;
}
i--;
nums.push(num);
break;
}
}
return nums.top();
}
private:
string getPostString(string s){
string postS = "";
stack<char> op;
int len = s.length();
for (int i = 0; i<len; i++){
switch (s[i]){
case ' ':
break;
case '(':
op.push('(');
break;
case ')':
while (!op.empty() && op.top() != '('){
postS += op.top();
op.pop();
}
if (!op.empty() && op.top() == '('){
op.pop();
}
break;
case '+':
case '-':
while (!op.empty() && op.top() != '('){
postS += op.top();
op.pop();
}
op.push(s[i]);
break;
default:
while (i<len && s[i] >= '0'&&s[i] <= '9'){
postS += s[i];
i++;
}
postS += '#'; //分隔数字
i--;
break;
}
}
while (!op.empty()){
postS += op.top();
op.pop();
}
return postS;
}
};转载请注明：康瑞部落 » [LeetCode] Basic Calculator
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