1190. 反转每对括号间的子串

/* 
a(bcdefghijkl(mno)p)q

a [‘a‘]
( [‘a‘, ‘‘]
b [‘a‘, ‘b‘]
c [‘a‘, ‘bc‘]
d [‘a‘, ‘bcd‘]
e [‘a‘, ‘bcde‘]
f [‘a‘, ‘bcdef‘]
g [‘a‘, ‘bcdefg‘]
h [‘a‘, ‘bcdefgh‘]
i [‘a‘, ‘bcdefghi‘]
j [‘a‘, ‘bcdefghij‘]
k [‘a‘, ‘bcdefghijk‘]
l [‘a‘, ‘bcdefghijkl‘]
( [‘a‘, ‘bcdefghijkl‘, ‘‘]
m [‘a‘, ‘bcdefghijkl‘, ‘m‘]
n [‘a‘, ‘bcdefghijkl‘, ‘mn‘]
o [‘a‘, ‘bcdefghijkl‘, ‘mno‘]
) [‘a‘, ‘bcdefghijklonm‘]
p [‘a‘, ‘bcdefghijklonmp‘]
) [‘apmnolkjihgfedcb‘]
q [‘apmnolkjihgfedcbq‘]

作者：tuotuoli
链接：https://leetcode-cn.com/problems/reverse-substrings-between-each-pair-of-parentheses/solution/1190-fan-zhuan-mei-dui-gua-hao-jian-de-zi-chuan-ji/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。*/

//思路
//temp为当前输入的连在一起的字符串
//若是字符 则放到temp后
//若遇到 左括号 将temp入栈 temp = ""
//若遇到 右括号 令temp = top + reverse(temp)
class Solution {
public:
    string reverseParentheses(string s) {
        stack<string> my_stack;
        string temp="";
        for (char ch : s)
        {
            if (ch == ‘(‘)
            {
                my_stack.push(temp);
                temp = "";
            }
            else if (ch == ‘)‘)
            {
                reverse(temp.begin(), temp.end());
                temp = my_stack.top() + temp;
                my_stack.pop();
            }
            else
                temp += ch;
        }
        return temp;
    }
};