## 题目描述：

LeetCode 443. String Compression

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.

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:

1. All characters have an ASCII value in `[35, 126]`.
2. `1 <= len(chars) <= 1000`.

## Python代码：

``````class Solution(object):
def compress(self, chars):
"""
:type chars: List[str]
:rtype: int
"""
last, n, y = chars, 1, 0
for x in range(1, len(chars)):
c = chars[x]
if c == last: n += 1
else:
for ch in last + str(n > 1 and n or ''):
chars[y] = ch
y += 1
last, n = c, 1
for ch in last + str(n > 1 and n or ''):
chars[y] = ch
y += 1
while len(chars) > y: chars.pop()
return y
``````

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