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.
当时做的时候没有考虑空间复杂度，直接模拟
class Solution {
public:
int compress(vector<char>& chars) {
int n = chars.size();
vector<char> tmp = chars;
chars.clear();
for (int i = 0; i < n; ++i) {
int x = 1;
while(i+1 < n && tmp[i] == tmp[i+1]) {
++i;
x++;
}
if (x == 1) {
chars.push_back(tmp[i]);
} else if (x >= 2) {
chars.push_back(tmp[i]);
vector<char> s;
while(x > 0) {
s.push_back(char(x%10 +'0'));
x /= 10;
}
for (int j = s.size()-1; j >= 0; --j) chars.push_back(s[j]);
}
}
return chars.size();
}
};