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443. String Compression

程序员文章站 2022-03-12 18:29:46
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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:

  1. All characters have an ASCII value in [35, 126].
  2. 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;
    }
}