# leetcode\_443

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. **two pointers**

```cpp
class Solution {
public:
    int compress(vector<char>& chars) {
        if (chars.size() <= 1) return chars.size();
        int w = 0;
        for (int i = 0, r = 0; r < chars.size(); r++) {
            if (r == chars.size() - 1 || chars[r] != chars[r + 1]) {
                chars[w++] = chars[i];
                if (r - i)
                    for (auto & c : to_string(r - i + 1))
                        chars[w++] = c;
                i = r + 1;
            }
        }

        return w;
    }
};
```