[LeetCode] 443. String Compression_Easy tag: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:

1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

这个题目不难, 就是edge case有点麻烦, 然后我们这里用当前和下一个char来判断, 会比较好, 如果count >1, 则需要进行相应的处理.

Code

class Solution:
    def codeString(self, chars):
        write, count = 0, 1
        for index, c in enumerate(chars):
            if index +1 == len(chars) or chars[index + 1] != c:
                chars[write] = c
                write += 1
                if count >1:
                    num = str(count)
                    for i in range(len(num)):
                        chars[write] = num[i]
                        write += 1
                count = 1
            else:
                count += 1
        return write