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.
解题要点：
双指针 i 和 j，i 在原点，由 j 来遍历，当 j 检测到重复的数结束后，先把 i 加入数组（从头开始）并把 j 转换为char，也加进数组。
class Solution {
public int compress(char[] chars) {
if(chars.length == 1) return 1;
int curr = 0, i = 0, j = 1;
int n = chars.length;
while(i < n){
while(j < n && chars[i] == chars[j]) j++;
chars[curr++] = chars[i];
if(j - i == 1){
i = j;
continue;
}
char[] counts = String.valueOf(j-i).toCharArray();
for(char c : counts) chars[curr++] = c;
i = j;
}
return curr;
}
}