題目如下：

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".

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Example 2:

Input: ["a"]Output: Return 1, and the first 1 characters of the input array should be: ["a"]Explanation: Nothing is replaced.

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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.

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Note:

All characters have an ASCII value in [35, 126].

1 <= len(chars) <= 1000.

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解題思路：從頭到尾遍歷數(shù)組，記錄連續(xù)字符的個數(shù)，然后插入數(shù)組前部，注意每次插入要記錄當(dāng)前的偏移量offset?。

代碼如下：

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

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轉(zhuǎn)載于:https://www.cnblogs.com/seyjs/p/9275563.html

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