443. String Compression

https://leetcode.com/problems/string-compression/

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.

Thoughts

将给定由[a-z]组成的string在原string上压缩：重复出现（频数>1）的字符替换成字符 + 频数。前后双指针，后指针每次前进一格并比较当前字符和后面是否一致（或到了最后一个字符），一致时频数+1，否则在前指针位置开始替换。

Code

class Solution:
    def compress(self, chars: List[str]) -> int:
        res, cnt, j = 0, 0, 0
        for i, c in enumerate(chars):
            cnt += 1
            if i == len(chars) - 1 or c != chars[i + 1]:   
                chars[j] = c
                j += 1
                if cnt != 1:
                    for k in str(cnt):
                        chars[j] = k
                        j += 1
                cnt = 0
        return j

class Solution {
    public int compress(char[] chars) {
        int pos = 0;
        for (int i = 0, count = 0; i < chars.length; i++) {
            count++;
            if (i == chars.length - 1 || chars[i] != chars[i + 1]) {
                chars[pos++] = chars[i];
                if (count != 1) {
                    char[] arr = String.valueOf(count).toCharArray();
                    for (int j = 0; j < arr.length; j++) {
                        chars[pos++] = arr[j];
                    }
                }
                count = 0;
            }
        }
        return pos;
    }
}

Analysis

时间复杂度O(N).