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].
    2. 1 <= len(chars) <= 1000.

    Runtime: 8 ms, faster than 42.86% of C++ online submissions for String Compression.

    1. class Solution {
    2. public:
    3.     int compress(vector<char>& chars) {
    4.         int size = chars.size();
    5.         if (size < 2) {
    6.             return size;
    7.         }
    8.         char current = chars[0];
    9.         int count = 1;
    10.         int j = 1;
    11.         for (int i = 1; i <= size; i++) {
    12.             if (i < size && current == chars[i]) {
    13.                 count++;
    14.                 continue;
    15.             }
    16.             chars[j - 1] = current;
    17.             if (i < size) {
    18.                 current = chars[i];
    19.             }
    20.             if (count == 1) {
    21.                 j++;
    22.             } else if (count < 10) {
    23.                 chars[j] = (char)(count + 48);
    24.                 j += 2;
    25.             } else {
    26.                 string s = to_string(count);
    27.                 for (char c: s) {
    28.                     chars[j++] = c;
    29.                 }
    30.                 j++;
    31.             }
    32.             count = 1;
    33.         }
    34.         return j - 1;
    35.     }
    36. };