题目要求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? 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.对字符串进行简单的压缩操作，压缩的规则是，如果出现多个重复的字母，则用字母加上字母出现的字数进行表示。如果字母只出现一次，则不记录次数。思路和代码核心思路是用三个指针分别记录三个下标：p1: 记录压缩后的内容的插入下标p2: 记录当前相同字符串的起始位置p3: 记录当前和起始位置比较的字符串的位置一旦出现p3的值不等于p2或是p3的值大于字符数组的长度，则将压缩结果从p1开始填写，实现O(1)的空间复杂度。 public int compress(char[] chars) { int p1 = 0; int p2 = 0; int p3 = 1; while(p2 < chars.length) { if(p3 >= chars.length || chars[p3] != chars[p2]) { int length = p3 - p2; chars[p1++] = chars[p2]; if(length != 1) { int count = 0; while(length != 0) { int num = length % 10; for(int i = p1+count ; i>p1 ; i–) { chars[i] = chars[i-1]; } chars[p1] = (char)(‘0’ + num); length /= 10; count++; } p1 += count; } p2 = p3; } p3++; } return p1; }