【leetcode】224. Basic Calculator
作者:互联网
Given a string
s
representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.
Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval()
.
Example 1:
Input: s = "1 + 1"
Output: 2
Example 2:
Input: s = " 2-1 + 2 "
Output: 3
Example 3:
Input: s = "(1+(4+5+2)-3)+(6+8)"
Output: 23
class Solution {
public:
int calculate(string s) {
//return (int) eval(s); 静态语言没有eval这种函数
// 因为 只有加减符号 而且都有效 括号直接差 没有优先级的差别吧 基本思路是对的 为啥写着写着 迷了
int res=0,sign=1,n=s.size();
stack<int> st;
for(int i=0;i<n;++i)
{
char c=s[i];
if(c>='0')
{
int num=0;
while(i<n &&s[i]>='0')
{
num=num*10+(s[i++]-'0');
}
res+=sign*num;
--i;
}
else if(c=='+')
{
sign=1;
}
else if(c=='-')
{
sign=-1;
}
else if(c=='(')
{
st.push(res);
st.push(sign);
res=0;
sign=1;
}
else if(c==')')
{
res*=st.top();st.pop();
res+=st.top();st.pop();
}
}
return res;
}
};
标签:int,res,Calculator,sign,else,num,st,224,leetcode 来源: https://www.cnblogs.com/aalan/p/15737736.html