题目地址：https://leetcode.com/problems/string-compression/description/open in new window
题目描述
Given an array of characters, compress it in-place.
Thelength 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、 AllcharactershaveanASCIIvaluein[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)
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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
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