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

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

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目录

• • 题目描述
  • 题目大意
  • 解题方法
  • • 使用额外空间
    • 不使用额外空间
  • 日期

题目地址：https://leetcode.com/problems/string-compression/description/

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:

1. All characters have an ASCII value in [35, 126].
2. 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