Alice is attempting to type a specific string on her computer. However, she tends to be clumsy and may press a key for too long, resulting in a character being typed multiple times.

Although Alice tried to focus on her typing, she is aware that she may still have done this at most once.

You are given a string word, which represents the final output displayed on Alice's screen.

Return the total number of possible original strings that Alice might have intended to type.

Example 1:
Input: word = "abbcccc"
Output: 5

Explanation:
The possible strings are: "abbcccc", "abbccc", "abbcc", "abbc", and "abcccc".

Example 2:
Input: word = "abcd"
Output: 1

Explanation:
The only possible string is "abcd".

Example 3:
Input: word = "aaaa"
Output: 4

Constraints:
1 <= word.length <= 100
word consists only of lowercase English letters.

找到初始輸入字符串 I。

Alice 正在她的電腦上輸入一個字符串。但是她打字技術比較笨拙，她 可能 在一個按鍵上按太久，導致一個字符被輸入 多次 。

盡管 Alice 盡可能集中注意力，她仍然可能會犯錯 至多 一次。

給你一個字符串 word ，它表示 最終 顯示在 Alice 顯示屏上的結果。

請你返回 Alice 一開始可能想要輸入字符串的總方案數。

思路

這道題不涉及算法，注意觀察字符串的規律。如果相鄰的字符相同，那么第二個字母就有可能是一次重復的輸入。所以我們可以遍歷 input 字符串，如果當前字母跟前一個字母不一樣，則說明沒有犯錯；如果跟前一個字母一樣，那么第二個字母就有可能是因為犯錯而輸入的，則計數加一。

復雜度

時間O(n)
空間O(1)

代碼

Java實現

class Solution {
    public int possibleStringCount(String word) {
        int count = 1;
        int n = word.length();
        for (int i = 1; i < n; i++) {
            if (word.charAt(i) == word.charAt(i - 1)) {
                count++;
            }
        }
        return count;
    }
}