原题链接在这里：

题目：

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.

题解：

计数连续相同char的个数加在后面.

Time Complexity: O(chars.length). Space: O(1).

AC Java:

1 class Solution { 2     public int compress(char[] chars) { 3         if(chars == null || chars.length == 0){ 4             return 0; 5         } 6          7         int mark = 0; 8         int write = 0; 9         for(int i = 0; i