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:
1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.
注意是in-place,不可再开辟存储空间，在chars内完成操作
class Solution {
public:
int compress(vector<char>& chars) {
if(chars.size()==0)
return 0;
int ans=0,index=0;
while(index<chars.size())
{
char c=chars[index];
int count=0;
while(index<chars.size() && chars[index]==c)
{
index++;
count++;
}
chars[ans++]=c;
if(count!=1)
{
for(char a:to_string(count))
chars[ans++]=a;
}
}
return ans;
}
};