443. String Compression -E

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:

1. All characters have an ASCII value in [35, 126].

A:

---------------------就是简单的数数----------

class Solution {
public:
    int compress(vector<char>& chars) {
        int n = chars.size();
        int curBegin=0, curEnd=0, nextWriteIndex = 0;
        while(curBegin<n)
        {
            char ch = chars[curBegin];
            while(curEnd+1 <n && chars[curEnd+1] == ch)
            {
                curEnd++;
            }
            int count = curEnd-curBegin+1;
            chars[nextWriteIndex++] = ch;
            if(count >1 )
            {
                for(auto c:to_string(count))
                {
                    chars[nextWriteIndex++] = c;
                }
            }
            curBegin = curEnd +1;
        }
        return nextWriteIndex;
    }
};

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