python-3.x - 检查字符串中连续出现了多少次[已解决]

Question

我想显示最常出现的数字或字母
在给定的字符串或数字或两者中连续
例子
s = 'aabskeeebadeeee'
输出: e 连续出现 4 次
我想过设置字符串然后为每个元素循环字符串
检查是否与元素集元素相等，如果是，则计数 =+1 并检查是否
在它旁边不等于将计数器值添加到具有相同索引的列表中
设置，如果值大于现有值，则将计数器值添加到 li 列表。
问题是错误索引超出或范围，尽管我想我正在看它。

s = 'aabskeeebadeeee'

c = 0
t = list(set(s)) # list of characters in s
li=[0,0,0,0,0,0]  # list for counted repeats
print(t)
for x in t:
    h = t.index(x)
    for index, i in enumerate(s):
        maximus = len(s)
        if i == x:
            c += 1
            if index < maximus:
                if s[index +1] != x:  # if next element is not x
                    if c > li[h]:   #update c if bigger than existing
                        li[h] = c
                    c = 0
            else:
                if c > li[h]:
                    li[h] = c


for i in t:
    n = t.index(i)
    print(i,li[n])

print(f'{s[li.index(max(li))]} appears {max(li)} consecutive times')  

已解决
我不想使用任何库，所以我又坐了下来，尽管它让我头疼，但我还是找到了解决方案:-)。

 s = 'aabskaaaabadcccc'


lil = tuple(set(s)) # Set a characters in s to remove duplicates and then make a tuple

li=[0,0,0,0,0,0]  # list for counted repeats, the index of number repeats for character
                # will be equal to index of its character in a tuple


for i in lil:    #iter over tuple of letters
    c = 0        #counter
    h= lil.index(i)  #take an index 
    for letter in s:            #iterate ove the string characters 
        if letter == i:         # check if equal with character from tuple
            c += 1              # if equal Counter +1
            if c > li[lil.index(letter)]:   # Updated the counter if present is bigger than the one stored.
                li[lil.index(letter)] = c
        else:
            c=0
            continue

m = max(li)

for index, j in enumerate(li):              #Check if the are characters with same max value
    if li[index] == m:
        print(f'{lil[index]} appears {m} consecutive times')

输出:

c appears 4 consecutive times
a appears 4 consecutive times

Answer 1

这是一个 O(n)时间，O(1)空间解决方案，通过返回先前看到的字符来打破联系:

def get_longest_consecutive_ch(s):
    count = max_count = 0
    longest_consecutive_ch = previous_ch = None
    for ch in s:
        if ch == previous_ch:
            count += 1
        else:
            previous_ch = ch
            count = 1
        if count > max_count:
            max_count = count
            longest_consecutive_ch = ch
    return longest_consecutive_ch, max_count

s = 'aabskeeebadeeee'
longest_consecutive_ch, count = get_longest_consecutive_ch(s)
print(f'{longest_consecutive_ch} appears {count} consecutive times in {s}')

输出:

e appears 4 consecutive times in aabskeeebadeeee