443 String Compression

本文介绍了一种在原地压缩字符数组的算法实现,该算法确保压缩后的数组长度不大于原数组,并且尽可能使用O(1)额外空间。通过几个实例展示了如何将重复字符替换为字符加计数的方式进行压缩。

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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.
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 && currChar == chars[index]) {
                index++;
                count++;
            }
            chars[indexAns++] = currChar;
            if (count > 1) {
                for (char c : Integer.toString(count).toCharArray()){
                    chars[indexAns++] = c;
                }
            }
        }
        return indexAns;
    }
}

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