https://leetcode-cn.com/problems/basic-calculator/
Idea: What a boring problem... Just pay attention to the details, and you can directly recurse.
class Solution {
public:
using pr=pair<int,int>;
pr dfs(const string &s,int idx)
{
int n=s.size(),sum=0,tmp=0,sign=1;
while(idx<n)
{
if(s[idx]=='(')
{
pr p=dfs(s,idx+1);
sum+=sign*p.first;
idx=p.second;
tmp=0,sign=1;
}
else if(s[idx]==')')
break;
else if(s[idx]==' ')
;
else if(s[idx]>='0'&&s[idx]<='9')
tmp=tmp*10+(s[idx]-'0');
else
{
sum+=sign*tmp;
tmp=0;
sign=s[idx]=='+'?1:-1;
}
++idx;
}
sum+=sign*tmp;
return pr(sum,idx);
}
int calculate(string s) {
return dfs(s,0).first;
}
};
Or use a stack. The idea is to store the current symbol: the positive sign is 1, and the negative sign is -1. To be precise, it maintains the symbol before the brackets, so that the final correct symbol can be directly calculated within the brackets (equivalent to putting The parentheses are removed and expanded).
class Solution {
public:
int calculate(string s) {
stack<int> signs;
int ans=0,sign=1,n=s.size(),tmp=0,i=0;
signs.push(1);
while(i<n)
{
if(s[i]==' ')
;
else if(s[i]=='(')
signs.push(sign);
else if(s[i]==')')
signs.pop();
else if(s[i]=='+')
sign=signs.top();
else if(s[i]=='-')
sign=-signs.top();
else
{
tmp=0;
while(i<n&&s[i]>='0'&&s[i]<='9')
tmp=tmp*10+(s[i]-'0'),++i;
ans+=sign*tmp;
--i;
}
++i;
}
return ans;
}
};