### 题目

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.

• 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.

### 用3个指针

[a, a, a, b, c, c, c, c, d, d]

start        end
|           |
[a, 3, b, b, c, c, c, c, d, d]
|
pen


#### 代码

class Solution {

public int compress(char[] chars) {
int pen = 0;
int start = 0;
while (start < chars.length) {
for (int end = start + 1; end <= chars.length; end++) {
if (end == chars.length || chars[end] != chars[start]) {
chars[pen++] = chars[start];
if (end - start > 1) {
String len = String.valueOf(end - start);
for (int i = 0; i < len.length() ; i++) {
chars[pen++] = len.charAt(i);
}
}
start = end;
break;
}
}
}
return pen;
}

}


#### 结果 