題目敘述
題目會給我們一個輸入陣列chars，要求我們把同樣的字元作分類並且依照長度做編碼壓縮。壓縮時，必須在陣列做in-place原地更新，最後回傳壓縮之後的字串長度。
壓縮規則:
同樣的字元一群，首項放字元，假如這群的長度只有1，那這群的壓縮就完成了。假如假如這群的長度大於1，那麼後面緊接著以字元的形式，放入這群的長度。
例如輸入['a','a','a'] 壓縮完變成['a','3'] 長度為2
例如輸入['a','b'] 壓縮完變成['a','b'] 長度為2
測試範例
Example 1:
Input: chars = ["a","a","b","b","c","c","c"]
Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
Example 2:
Input: chars = ["a"]
Output: Return 1, and the first character of the input array should be: ["a"]
Explanation: The only group is "a", which remains uncompressed since it's a single character.
Example 3:
Input: chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
Output: Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
Explanation: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".
約束條件
1 <= chars.length <= 2000
輸入陣列chars的長度介於1~2000之間
chars[i]is a lowercase English letter, uppercase English letter, digit, or symbol.
chars內不包含的只會是大寫英文字母、小寫英文字母、數字、或者符號。
演算法
這邊可以使用經典的雙指針來解。
宣告兩個變數，slow和fast，初始化為零，代表起頭位置。
從左邊開始向右掃描每一群，slow負責壓縮後對應要填的位置，fast負責去量測同一群字元的長度，並且紀錄在計數器count裡面。
程式碼
class Solution:
def compress(self, chars: List[str]) -> int:
slow, fast = 0, 0
n = len(chars)
while fast < n:
# Update leading character
chars[slow] = chars[fast]
count = 1
# Measure the length of cluster with same character
while (fast + 1) < n and chars[fast] == chars[fast+1]:
fast += 1
count += 1
# Write length of current cluster
if count > 1:
for c in str(count):
chars[slow+1] = c
slow += 1
# Moving to right hand side
fast += 1
slow += 1
return slow
