Leetcode 443 String Compression

基础的字符串操作,用两个指针扫一遍就行.

class Solution(object):
    def compress(self,chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        if not chars:
            return 0
        if len(chars) == 1:
            return 1
        left, right, num = 0, 1, 1
        while right <= len(chars):
            if right == len(chars):
                if num > 1:
                    chars[left:right] = chars[left] + str(num)
                break
            if chars[right] == chars[left]:
                num += 1
                right += 1
            elif num > 1:  # need modification
                chars[left:right] = chars[left] + str(num)
                left = left + len(chars[left] + str(num))
                right = left + 1
                num = 1
            else:
                left = right
                right = left + 1
        print(chars)
        return len(chars)

原文地址：https://www.cnblogs.com/zywscq/p/10504034.html