Given a?0-indexed?string?word?and a character?ch,?reverse?the segment of?word?that starts at index?0?and ends at the index of the?first occurrence?of?ch?(inclusive). If the character?ch?does not exist in?word, do nothing.

• For example, if?word = "abcdefd"?and?ch = "d", then you should?reverse?the segment that starts at?0?and ends at?3?(inclusive). The resulting string will be?"dcbaefd".

Return?the resulting string.

?

Example 1:

Input: word = "abcdefd", ch = "d"

Output: "dcbaefd"

Explanation:?The first occurrence of "d" is at index 3. Reverse the part of word from 0 to 3 (inclusive), the resulting string is "dcbaefd".

Example 2:

Input: word = "xyxzxe", ch = "z"

Output: "zxyxxe"

Explanation:?The first and only occurrence of "z" is at index 3. Reverse the part of word from 0 to 3 (inclusive), the resulting string is "zxyxxe".

Example 3:

Input: word = "abcd", ch = "z"

Output: "abcd"

Explanation:?"z" does not exist in word. You should not do any reverse operation, the resulting string is "abcd".

?stringbuilder 有自定義的reverse()函數(shù)，直接用就行，就不需要自己一個一個去遍歷了

easy題目，直接按照題意模擬即可；

下面是代碼：

Constraints:

• 1 <= word.length <= 250

• word?consists of lowercase English letters.

• ch?is a lowercase English letter.

Runtime:?1 ms, faster than?47.79%?of?Java?online submissions for?Reverse Prefix of Word.

Memory Usage:?40.6 MB, less than?87.13%?of?Java?online submissions for?Reverse Prefix of Word.