Description

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?

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".

即对Char型数组中的字符进行统计，要求原地修改原数组，要注意原始字符的顺序位置

Solution

双指针

统计字符，首先想到用 HashMap，但是原题中要求是对连续字符的统计，而不是对整个数组中相同字符的统计，所以 HashMap 并不是好的选择；
使用双指针统计个数，anchor 指向当前字符，read遍历数组，判断相邻两个字符是否相同；
write 指针用来修改数组，当 read 判断到相邻两个字符不相同时，对原数组修改，连续字符个数为 read - anchor +1；
返回修改后数组长度 → 最后write指针指向的index值;
需要注意当某字符个数不是个位数时，需要用char类型分开存储，方法为 ( int + "" ).toCharArray();

class Solution {
    public int compress(char[] chars) {
        int write  = 0, anchor = 0;
        for(int read = 0;read < chars.length;read++){
            if(read + 1 == chars.length || chars[read + 1] != chars[read]){  //对 read + 1 == chars.length 的判断需要放前面，不然会溢出
                chars[write++] = chars[anchor];
                if(read > anchor){
                    for(char c : ("" + (read - anchor +1)).toCharArray()){
                         chars[write++] = c;
                    }
                }
                anchor = read + 1;
            }
        }
        return write;
    }
}

解法来源