443. String Compression
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.
class Solution:
def compress(self, chars):
"""
:type chars: List[str]
:rtype: int
"""
pos = 0
while pos<len(chars):
temp = chars[pos]
count = 1
while pos+1<len(chars) and chars[pos+1]==temp:
count += 1
chars.pop(pos+1)
if count>1:
for i in reversed(str(count)):
chars.insert(pos+1,i)
pos += 2
else:
pos += 1
# print(chars)
return len(chars)
