Given an array of characters, compress it in-place.

给定一组字符，就地压缩它。

The length after compression must always be smaller than or equal to the original array.

压缩后的长度必须始终小于或等于原始数组。

Every element of the array should be a character (not int) of length 1.

数组的每个元素都应该是长度为1的字符（不是int）。

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

在就地修改输入数组后，返回数组的新长度。

Follow up:

跟进：
Could you solve it using only O(1) extra space?

你能用O（1）额外空间解决它吗？

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"]返回6，输入数组的前6个字符应为：[“a”，“2”，“b”，“2”，“c”，“3”]Explanation:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".“aa”被“a2”取代。 “bb”被“b2”取代。 “ccc”被“c3”取代。

Example 2:

Input:
["a"]Output:
Return 1, and the first 1 characters of the input array should be: ["a"]返回1，输入数组的前1个字符应为：[“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"].返回4，输入数组的前4个字符应为：[“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.由于字符“a”不重复，因此不会压缩。 “bbbbbbbbbbbbb”被“b12”取代。请注意，每个数字在数组中都有自己的条目。

Note:

1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

 1 class Solution {
 2     public int compress(char[] chars) {
 3         int slow=0,fast=0;
 4         while(fast<chars.length){
 5             char currChar=chars[fast];
 6             int count=0;
 7             while(fast<chars.length && chars[fast]==currChar){
 8                 fast++;
 9                 count++;
10             }
11             chars[slow++]=currChar;
12             if(count!=1)
13                 for(char c:Integer.toString(count).toCharArray())
14                     chars[slow++]=c;
15         }
16         return slow;
17     }
18 }