1. 题意

当字符串出现$i$时，反转 当前出现的字符串。

故障键盘

2. 题解

• 直接模拟即可
  复杂度$o(n^2)$

class Solution {
public:string finalString(string s) {string ans;for (auto c:s) {if (c == 'i') {reverse(ans.begin(), ans.end());}else {ans.push_back(c);}}return ans;}
};

• 双端队列
  当出现翻转字符的时候，换方向进行插入

class Solution {
public:string finalString(string s) {deque<char> q;bool back_insert = true;for (auto &c:s) {if (c == 'i') back_insert = !back_insert;else if (back_insert)q.push_back(c);elseq.push_front(c);}return back_insert ? string(q.begin(), q.end()):string(q.rbegin(), q.rend());}
};