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.

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.

方法1: two pointers

思路：

用双指针的方法，慢指针指向下一个chars需要被改写的位置，j向前寻找新的字母，并且累计一个count来记录有多少个重复出现过。当统计结束时，此时j指向一个新的字母，可以开始写入刚刚结束的这个字符。首先写入字符本身，慢指针前进一步，然后判断是否count == 1，当不为1的时候才需要写入重复次数。注意12是被拆分成“1”， “2”写入的，所以要对count进行一下to_string的转换并遍历。

Complexity

Time complexity: O(n)
Space complexity: O(1)

class Solution {
public:
    int compress(vector<char>& chars) {
        int i = 0, j = 0, n = chars.size();
        while (j < n) {
            char current = chars[j];
            int count = 0;
            while (j < n && chars[j] == current) {
                j++;
                count++;
            }
            chars[i++] = current; 
            
            if (count != 1) {
                for (auto c: to_string(count)) {
                    chars[i++] = c;
                }
            }
        }
        return i;
    }
};