leetcode443. String Compression

Posted 2020-12-02 seyjs

题目如下：

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:

1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

 

解题思路：从头到尾遍历数组，记录连续字符的个数，然后插入数组前部，注意每次插入要记录当前的偏移量offset 。

代码如下：

class Solution(object):
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        lastChar = None
        count = 0
        inx = 0
        offset = 0
        chars.append(‘END‘) # terminator
        while inx < len(chars):
            i = chars[inx]
            if lastChar == None:
                lastChar = i
                count = 1
            elif lastChar == i:
                count += 1
            else:
                lastOff = offset
                chars.insert(offset,lastChar)
                offset += 1
                if count != 1:
                    count = str(count)
                    for j in count:
                        chars.insert(offset, j)
                        offset += 1
                lastChar = i
                count = 1
                inx += (offset - lastOff)
            inx += 1
        #print chars
        del chars[-1]
        return offset