标签:
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:
Note: Do not use the eval built-in library function.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
和Basic Calculator比較相似。都是很巧妙的解法。
runtime:24ms
class Solution {
public:
int calculate(string s) {
int len=s.size();
if(s.empty())
return 0;
stack<int> st;
int num = 0;
char sign = ‘+‘;
for(int i=0;i<len;i++){
if(isdigit(s[i])){
num = num*10+s[i]-‘0‘;
}
if(!isdigit(s[i]) &&‘ ‘!=s[i] || i==len-1){
if(sign==‘-‘){
st.push(-num);
}
if(sign==‘+‘){
st.push(num);
}
if(sign==‘*‘){
int tmp=st.top();
st.pop();
st.push(tmp*num);
}
if(sign==‘/‘){
int tmp=st.top();
st.pop();
st.push(tmp/num);
}
sign = s[i];
num = 0;
}
}
int re = 0;
while(!st.empty()){
re += st.top();
st.pop();
}
return re;
}
};
LeetCode227:Basic Calculator II
标签:
原文地址:http://www.cnblogs.com/bhlsheji/p/5314037.html