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class Count{
public int Count(String a){
int n = a.length();
int result = 0;
int num = 0;
int p = 0;
int L1 = a.indexOf(‘(‘);
int R1 = a.indexOf(‘)‘);
String b = a.substring(L1+1,R1-1);
int op = 0;
if(L1 >= 0 ) return Count(b) + result;
else
{
for(int i=0 ;i < n ;i++)
{
if(‘0‘<(a.charAt(i))&& (a.charAt(i))<‘9‘) num = num*10+num;
switch(op)
{
case 0 : break;
case 1 : num = num + p ;break;
case 2 : num = p - num ;break;
default : break;
}
if(a.charAt(i)==‘+‘)
op = 1;
else if (a.charAt(i)==‘-‘)
op = 2 ;
else
op = 0;
p = num ;
}
result = num;
return result;
}
}
}这个鸟玩意用了迭代,其实不太好浪费空间,数据多了还容易死,有人用stack还没研究
这个是600mm的还没看明白
https://leetcode.com/discuss/39454/accepted-infix-postfix-based-solution-explaination-600ms
Java Solution (Basic Calculator) 还没测,晚上回去册
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原文地址:http://www.cnblogs.com/puck/p/4563729.html