227. Basic Calculator II
Implement a basic calculator to evaluate a simple expression string.
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:
"3+2*2" = 7 " 3/2 " = 1 " 3+5 / 2 " = 5
Note: Do not use the eval
思路：
用一个栈来保存数字，
如果之前的运算符是+，则将该数压入栈；
如果之前的运算符是-，则将该数的负数压入栈；
如果之前的运算符是*，则弹出栈顶数，和当前的数相乘压入栈；
如果之前的运算符是-，则弹出栈顶数，和当前的数相除压入栈；
注意处理最后一次的情况
class Solution {
public:
int calculate(string s) {
int len = s.size();
stack<int> stk;
int num = 0;
char sign = '+';
for (int i = 0; i < len; i++){
if (isdigit(s[i])){
num = num * 10 + s[i] - '0';
}
if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/'||i==len-1){//不加else
if (sign == '+'){
stk.push(num);
}
else if (sign == '-'){
stk.push(-num);
}
else if (sign == '*'){
int top = stk.top();
stk.pop();
stk.push(top*num);
}
else{
int top = stk.top();
stk.pop();
stk.push(top/num);
}
sign = s[i];
num = 0;
}
}
int res = 0;
while (!stk.empty()){
res += stk.top();
stk.pop();
}
return res;
}
};
224. Basic Calculator
Implement a basic calculator to evaluate a simple expression string.
( 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
遇到 '(' 就把之前的结果和符号push进stack. 遇到')'就把 当前结果*stack中的符号 再加上stack中之前的结果.
class Solution {
public:
int calculate(string s) {
stack<int> stk;
int len = s.size();
int res = 0;
int sign = 1;
int i = 0;
while (i<len){
if (isdigit(s[i])){
int num = 0;
while (i < len&&isdigit(s[i])){
num = num * 10 + s[i] - '0';
i++;
}
res += num*sign;
}
else if (s[i] == '+'){
sign = 1;
i++;
}
else if (s[i] == '-'){
sign = -1;
i++;
}
else if (s[i] == '('){
stk.push(res);
stk.push(sign);
res = 0;
sign = 1;
i++;
}
else if (s[i] == ')'){
res = res*stk.top();
stk.pop();
res += stk.top();
stk.pop();
i++;
}
else{
i++;
}
}
return res;
}
};
