【LeetCode】443. String Compression 解题报告（Python）

作者：负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/

题目描述

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:

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

题目大意

统计每个字符出现的次数，然后放到原地，需要按照顺序放。完成了字符串的压缩。

解题方法

使用额外空间

自己的方法比较简单粗暴，用了额外的空间来保存了所有的数字出现的次数，最后再放回到chars上。

class Solution(object):

    def compress(self, chars):

        """

        :type chars: List[str]

        :rtype: int

        """

        marks = ""

        length = -1

        cur = chars[0]

        for i, value in enumerate(chars):

            length += 1

            if value != cur:

                count = str(length) if length != 1 else ''

                marks += cur + count

                cur = value

                length = 0

            if i == len(chars) - 1:

                length += 1

                count = str(length) if length != 1 else ''

                marks += cur + count

                cur = value

                length = 0

        print marks

        for i, mark in enumerate(marks):

            chars[i] = mark

        return len(marks)

不使用额外空间

保存一个pos位置，告诉我们当前需要放在哪个地方。然后我们统计连续的字符出现了多少次，如果大于1次才往后拼接上去。

class Solution(object):

    def compress(self, chars):

        """

        :type chars: List[str]

        :rtype: int

        """

        pre = chars[0]

        count = 0

        pos = 0

        for ch in chars:

            if pre == ch:

                count += 1

            else:

                chars[pos] = pre

                pos += 1

                if count > 1:

                    count = str(count)

                    for i in range(len(count)):

                        chars[pos] = count[i]

                        pos += 1

                count = 1

                pre = ch

        chars[pos] = pre

        pos += 1

        if count > 1:

            count = str(count)

            for i in range(len(count)):

                chars[pos] = count[i]

                pos += 1

        return pos

日期

2018 年 1 月 27 日
2018 年 11 月 24 日 —— 周六快乐

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