文巾解题 1446. 连续字符

很简单，从头到尾一次遍历即可，时间复杂度O(n)，空间复杂度O(1)

class Solution:
    def maxPower(self, s: str) -> int:
        i=0
        num=1
        max_num=0
        c=''
        for i in s:
            if(i==c):
                num+=1
            else:
                c=i
                num=1
            if(num>max_num):
                max_num=num
        return(max_num)