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.
双指针+一个常数用来计数
时间：O(N)，空间：O(1)
class Solution { public int compress(char[] chars) { int resIdx = 0, idx = 0; while(idx < chars.length) { char cur = chars[idx]; int cnt = 0; while(idx < chars.length && cur == chars[idx]) { idx++; cnt++; } chars[resIdx++] = cur; if(cnt != 1) { for(char c : Integer.toString(cnt).toCharArray()) { chars[resIdx++] = c; } } } return resIdx; }}