[LeetCode] 443. String Compression_Easy tag:String
2018-08-13 05:22 Johnson_强生仔仔 阅读(147) 评论(0) 编辑 收藏 举报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
.
这个题目不难, 就是edge case有点麻烦, 然后我们这里用当前和下一个char来判断, 会比较好, 如果count >1, 则需要进行相应的处理.
Code
class Solution: def codeString(self, chars): write, count = 0, 1 for index, c in enumerate(chars): if index +1 == len(chars) or chars[index + 1] != c: chars[write] = c write += 1 if count >1: num = str(count) for i in range(len(num)): chars[write] = num[i] write += 1 count = 1 else: count += 1 return write