# Solution1 双指针模拟

* 时间复杂度 $O(n)$
* 空间复杂度 $O(1)$

注意处理好特情况：

* size == 1: 直接返回 1
* size > 1

    遍历 (1, N)

  * 对中间的元素：
    * i != i-1，则要加，注意区分 cnt == 1
    * i == i-1，直接 cnt++
  * 对最后一个元素，要判断倒数第二和最后一个字符
    * 最后两个字符相同：直接按 cnt + 1 算
    * 最后两个字符不同：分别算 倒数第二个字符的 cnt，和最后一个字符


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