443. 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.
Note:
- All characters have an ASCII value in
[35, 126]
. -
1 <= len(chars) <= 1000
.
Solution
Two-pointer, time O(n), space O(1)
class Solution {
public int compress(char[] chars) {
int start = 0;
int count = 0;
for (int end = 0; end < chars.length; ++end) {
++count;
// meet the end of a squence of same chars
if (end == chars.length - 1 || chars[end] != chars[end + 1]) {
chars[start] = chars[end];
++start;
// copy count to start
if (count != 1) {
char[] arr = String.valueOf(count).toCharArray();
for (int i = 0; i < arr.length; ++i, ++start) {
chars[start] = arr[i];
}
}
count = 0;
}
}
return start;
}
}
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