443. 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.

class Solution:
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        pos = 0
        while pos<len(chars):
            temp = chars[pos]
            count = 1
            while pos+1<len(chars) and chars[pos+1]==temp:
                count += 1
                chars.pop(pos+1)
            if count>1:
                for i in reversed(str(count)):
                    chars.insert(pos+1,i)
                pos += 2
            else:
                pos += 1
        # print(chars)
        return len(chars)

443. String Compression

原文：https://www.cnblogs.com/bernieloveslife/p/10235856.html