You are given a string?s?of lowercase English letters and an integer array?shifts?of the same length.

Call the?shift()?of a letter, the next letter in the alphabet, (wrapping around so that?'z'?becomes?'a').

• For example,?shift('a') = 'b',?shift('t') = 'u', and?shift('z') = 'a'.

Now for each?shifts[i] = x, we want to shift the first?i + 1?letters of?s,?x?times.

Return?the final string after all such shifts to s are applied.

?

Example 1:

Input: s = "abc", shifts = [3,5,9]

Output: "rpl"

Explanation: We start with "abc".

After shifting the first 1 letters of s by 3,?

we have "dbc".

After shifting the first 2 letters of s by 5,?

we have "igc".?

After shifting the first 3 letters of s by 9,?

we have "rpl",

the answer.

Example 2:

Input: s = "aaa", shifts = [1,2,3]

Output: "gfd"

?

Constraints:

• 1 <= s.length <= 105

• s?consists of lowercase English letters.

• shifts.length == s.length

• 0 <= shifts[i] <= 109

• 一開始卡在char +數(shù)字那里了，就是加個括號（char）強(qiáng)制轉(zhuǎn)換一下即可。

• 然后提交結(jié)果報錯，看了下錯誤的case，又是很大的數(shù)字，，，呃，于是在+shifts[i]后面增加了%26.然后就直接過了，int這個類型真的是加減乘除都要考慮溢出了。。。。

Runtime:?13 ms, faster than?71.40%?of?Java?online submissions for?Shifting Letters.

Memory Usage:?52.9 MB, less than?56.27%?of?Java?online submissions for?Shifting Letters.