[leetcode]443. String Compression

链接：https://leetcode.com/problems/string-compression/description/

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.

 

class Solution {
public:
    int compress(vector<char>& chars) {
	int mark = 0;
	int i = 0;
	int count = 0;
	while (i < (int)chars.size()){
		if (chars[i] != chars[i + 1]||i == (int)chars.size() - 1 ){
			count++;
			chars[mark++] = chars[i];
			if (count>1){
				string temp = to_string(count);
				for (char c : temp){
					chars[mark++] = c;
				}
			}
			count = 0;
		}
		else{
			count++;
		}
		i++;
	}
	return mark;
}
};