题目如下：
Given an array of characters, compress it .
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 , 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.
解题思路：从头到尾遍历数组，记录连续字符的个数，然后插入数组前部，注意每次插入要记录当前的偏移量offset 。
代码如下：
class Solution(object): def compress(self, chars): """ :type chars: List[str] :rtype: int """ lastChar = None count = 0 inx = 0 offset = 0 chars.append('END') # terminator while inx < len(chars): i = chars[inx] if lastChar == None: lastChar = i count = 1 elif lastChar == i: count += 1 else: lastOff = offset chars.insert(offset,lastChar) offset += 1 if count != 1: count = str(count) for j in count: chars.insert(offset, j) offset += 1 lastChar = i count = 1 inx += (offset - lastOff) inx += 1 #print chars del chars[-1] return offset