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.
压缩字符串
C++（6ms）：
1 class Solution { 2 public: 3 int compress(vector<char>& chars) { 4 int len = chars.size() ; 5 if (len < 2) 6 return len ; 7 int res = 0 ; 8 char c = chars[0] ; 9 int num = 1 ; 10 chars.push_back(' ') ; 11 for(int i = 1 ; i < len+1 ; i++){ 12 if (chars[i] == chars[i-1]) 13 num++ ; 14 if (chars[i] != chars[i-1] ){ 15 chars[res++] = c ; 16 if (num > 1){ 17 string s = to_string(num) ; 18 for(int j = 0 ; j < s.size() ; j++){ 19 chars[res++] = s[j] ; 20 } 21 } 22 num = 1 ; 23 c = chars[i] ; 24 } 25 } 26 return res ; 27 } 28 };