## Description

Given an array of characters, compress it **in-place**.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a **character** (not int) of length 1.

After you are done **modifying the input array in-place**, return the new length of the array.

**Follow up:**
Could you solve it using only O(1) extra space?

**Example 1:**

```
Input:
["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:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".
```

**Example 2:**

```
Input:
["a"]
Output:
Return 1, and the first 1 characters of the input array should be: ["a"]
Explanation:
Nothing is replaced.
```

**Example 3:**

```
Input:
["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:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.
```

**Note:**

- All characters have an ASCII value in
`[35, 126]`

. `1 <= len(chars) <= 1000`

.

## Solutions

### 1. 不优雅的拼凑

自己实现的版本真的是一堆 if-else，说明没有对规律了解清楚：

```
# Time Complexity: O(n)
# Space Complexity: O(1)
class Solution:
def compress(self, chars: List[str]) -> int:
if not chars:
return 0
if len(chars) == 1:
return 1
n = len(chars)
pre = chars[0]
cnt = 1
idx = 0
res = ''
for i in range(1, n):
if chars[i] == pre:
cnt += 1
else:
chars[idx] = pre
pre = chars[i]
if cnt == 1:
idx += 1
elif cnt >= 10:
for c in str(cnt):
idx += 1
chars[idx] = c
idx += 1
else:
chars[idx+1] = str(cnt)
idx += 2
cnt = 1
if i == n-1:
chars[idx] = pre
if cnt == 1:
idx += 1
elif cnt >= 10:
for c in str(cnt):
idx += 1
chars[idx] = c
idx += 1
else:
chars[idx+1] = str(cnt)
idx += 2
return idx
# Runtime: 56 ms, faster than 98.75% of Python3 online submissions for String Compression.
# Memory Usage: 12.9 MB, less than 100.00% of Python3 online submissions for String Compression.
```

### 2. 优雅的版本

```
# Time Complexity: O(n)
# Space Complexity: O(1)
class Solution:
def compress(self, chars: List[str]) -> int:
n = len(chars)
i, count = 0, 1
for j in range(1,n+1):
if j < n and chars[j] == chars[j-1]:
count += 1
else:
chars[i] = chars[j-1]
i += 1
if count > 1:
for m in str(count):
chars[i] = m
i += 1
count = 1
return i
# Runtime: 60 ms, faster than 95.41% of Python3 online submissions for String Compression.
# Memory Usage: 12.8 MB, less than 100.00% of Python3 online submissions for String Compression.
```