Implement a basic calculator to evaluate a simple expression string.
The expression string may contain open (
and closing parentheses )
, the plus +
or minus sign -
, non-negative integers and empty spaces .
Example 1:
Input: "1 + 1" Output: 2
Example 2:
Input: " 2-1 + 2 " Output: 3
Example 3:
Input: "(1+(4+5+2)-3)+(6+8)" Output: 23
Note:
- You may assume that the given expression is always valid.
- Do not use the
eval
built-in library function.
思路:
其实这道题最重要的一点就是需要处理好括号的问题,我们这里使用stack来处理括号的优先级问题,这里的处理方式很像第32题的处理。
解法1:
我们先处理输入中的括号,然后处理完括号后再处理比较简单的情况。
class Solution {
public:
int calculate(string s) {
string num;
stack<string> ss;
for (char a : s) {
if (a == ' ') {
if (num == "") continue;
ss.push(num);
num = "";
continue;
}
else if (a == ')') {
if (num != "") {
ss.push(num);
num = "";
}
list<string> inBracket;
while (ss.top() != "(") {
inBracket.push_front(ss.top());
ss.pop();
}
ss.pop();
int sum = calList(inBracket);
ss.push(to_string(sum));
}
else if (a == '(' || a == '+' || a == '-') {
if (num != "") {
ss.push(num);
num = "";
}
ss.push(string(1,a));
}
else num += a;
}
if (num != "") {
ss.push(num);
num = "";
}
list<string> left;
while (!ss.empty()) {
left.push_front(ss.top());
ss.pop();
}
int sum = calList(left);
return sum;
}
private:
int calList(list<string> l) {
bool positive = true;
int sum = 0;
for (string s : l) {
if (s == "+") positive = true;
else if (s == "-") positive = false;
else {
int num = stoi(s);
if (positive) sum += num;
else sum -= num;
}
}
return sum;
}
};
解法2:
我们这里使用两个stack来分别存放数字和符号,代码如下:
class Solution {
public:
int calculate(string s) {
stack<int> nums, opt;
int sum = 0, sign = 1;
for (int i = 0; i<int(s.size()); i++) {
if (isdigit(s[i])) {
string numS;
while (isdigit(s[i]))
numS += s[i++];
i--;
sum += sign * stoi(numS);
}
else if (s[i] == '(') {
nums.push(sum);
opt.push(sign);
sum = 0; sign = 1;
}
else if (s[i] == '+') sign = 1;
else if (s[i] == '-') sign = -1;
else if (s[i] == ')') {
sign = opt.top(); opt.pop();
sum = sign * sum + nums.top(); nums.pop();
}
}
return sum;
}
};