LeetCode 443 String Compression

LeetCode

Given an array of characters chars, compress it using the following algorithm:

Begin with an empty string s. For each group of consecutive repeating characters in chars:

If the group’s length is 1, append the character to s. Otherwise, append the character followed by the group’s length.

The compressed string s should not be returned separately, but instead, be stored in the input character array chars. Note that group lengths that are 10 or longer will be split into multiple characters in chars.

After you are done modifying the input array, return the new length of the array.

You must write an algorithm that uses only constant extra space.

Example 1:

Input: chars = [“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: The groups are “aa”, “bb”, and “ccc”. This compresses to “a2b2c3”.

Example 2:

Input: chars = [“a”]
Output: Return 1, and the first character of the input array should be: [“a”]
Explanation: The only group is “a”, which remains uncompressed since it’s a single character.

Example 3:

Input: chars = [“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: The groups are “a” and “bbbbbbbbbbbb”. This compresses to “ab12”.

Example 4:

Input: chars = [“a”,”a”,”a”,”b”,”b”,”a”,”a”]
Output: Return 6, and the first 6 characters of the input array should be: [“a”,”3”,”b”,”2”,”a”,”2”].
Explanation: The groups are “aaa”, “bb”, and “aa”. This compresses to “a3b2a2”. Note that each group is independent even if two groups have the same character.

Constraints:

1 <= chars.length <= 2000 chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.

题目大意：

相邻相同字母用数字压缩

解题思路：

N/A

解题步骤：

N/A

注意事项：

题目要求，如果是超过10，也要将这个数按多个字符populate到原数组，见populate_count的实现，用字符串处理
在循环外处理最后一部分

Python代码：

def compress(self, chars: List[str]) -> int:
	res, count = 1, 1
	for i in range(1, len(chars)):
		if chars[i - 1] == chars[i]:
			count += 1
		else:
			if count > 1:
				res = self.populate_count(chars, res, count)
			count = 1
			chars[res] = chars[i]
			res += 1
	if count > 1:
		res = self.populate_count(chars, res, count)
	return res

def populate_count(self, chars, res, count):
	num_str = str(count)
	chars[res:res + len(num_str)] = [c for c in num_str]
	res += len(num_str)
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(1)