难度:Easy
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.
原地修改，需要注意重复元素出现次数，如果是1可以省略，后面补上的长度是次数的宽度。
class Solution {
public:
int compress(vector<char>& chars) {
int start=1;
int count=1;
int res=0;
for(int i=1;i<chars.size();i++)
{
if(chars[i]==chars[i-1])
count++;
else
{
if(count==1){
res+=1;
}
else
{
string tmp=to_string(count);
res+=1+tmp.length();
for(auto c:tmp)
chars[start++]=c;
count=1;
}
chars[start++]=chars[i];
}
}
if(count==1){
res+=1;
}
else
{
string tmp=to_string(count);
res+=1+tmp.length();
for(auto c:tmp)
chars[start++]=c;
}
return res;
}
};
执行用时 :12 ms, 在所有 C++ 提交中击败了89.58%的用户
内存消耗 :9.1 MB, 在所有 C++ 提交中击败了91.06%的用户