443. String Compression
Tags: ‘String’
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
O(n) O(1) 97% 6%
public int compress(char[] chars) {
int i = 0, end = 0;
while (i < chars.length) {
int count = 0;
char curr = chars[i];
while (i < chars.length && chars[i] ==curr) {
count++;
i++;
}
chars[end++] = curr;
if (count != 1) {
for (char c: Integer.toString(count).toCharArray()) {
chars[end++] = c;
}
}
}
return end;
}