原题
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.
解法
双指针法, 使用while遍历字符串, 用curr, next代表当前index和下一个index, 当两个字符相同时, 持续计数, 最后将结果放到new中. Edge case是最后一个字符是单个字符, 那么我们检查curr是否走到最后一个index, 如果是, 那么new加上最后一个字符. 最后深度复制new到chars. 返回new的长度
Time: O(n)
Space: O(1)
代码
class Solution(object):
def compress(self, chars):
"""
:type chars: List[str]
:rtype: int
"""
# base case
if len(chars) == 1:
return 1
new = []
curr, next = 0, 1
while next < len(chars):
count = 1
while next < len(chars) and chars[next] == chars[curr]:
count += 1
next += 1
if count == 1:
new.append(chars[curr])
else:
new.append(chars[curr])
new.extend(list(str(count)))
curr = next
next += 1
if curr == len(chars)-1:
new.append(chars[curr])
chars[:] = new
return len(new)