1446. 连续字符 思路:模拟 保存 m a x n maxn maxn记录最长连续字符 class Solution { public: int maxPower(string s) { int res = 1, maxn = 1; for (int i = 1; i < s.size(); i ++ ) { if (s[i] == s[i - 1]) res ++ , maxn = max(maxn, res); else res = 1; } return maxn; } };