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.
Solution:
class Solution {
public int compress(char[] chars) {
int index = 0, indexAns = 0;
while (index < chars.length) {
char currChar = chars[index];
int count = 0;
while (index < chars.length && chars[index] == currChar) {
count ++;
index ++;
}
chars[indexAns++] = currChar;
if (count > 1) {
for (char c : String.valueOf(count).toCharArray()) {
chars[indexAns++] = c;
}
}
}
return indexAns;
}
}