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# 1446.连续字符

[原题](https://leetcode-cn.com/problems/consecutive-characters/)

## 一、遍历

一次遍历，用一个计数器cnt记录当前连续字符长度，每次`s[i]` 与 `s[i-1]`比较，若相等则cnt++，否则cnt置1。每次得到新的cnt更新全局最大长度res

```python
class Solution:
    def maxPower(self, s: str) -> int:
        n = len(s)
        if n == 1:
            return 1
        cnt = 1
        res = 0
        for i in range(1, n):
            if s[i-1] == s[i] :
                cnt += 1
            else:
                cnt = 1
            res = max(res, cnt)
        return res
```