You are given a string?s?of lowercase English letters and a 2D integer array?shifts?where?shifts[i] = [starti, endi, directioni].?

For every?i,?shift?the characters in?s?from the index?starti?to the index?endi

?(inclusive) forward if?directioni?= 1,?

or shift the characters backward if?directioni?= 0.

Shifting a character?forward?means replacing it with the?next?letter in the alphabet

(wrapping around so that?'z'?becomes?'a').?

Similarly, shifting a character?backward?means replacing it with the?previous?letter in the alphabet (wrapping around so that?'a'?becomes?'z').

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

?

Example 1:

Input: s = "abc", shifts = [[0,1,0],[1,2,1],[0,2,1]]

Output: "ace"

Explanation: Firstly, shift the characters from index 0 to index 1 backward.?

Now s = "zac". Secondly, shift the characters from index 1 to index 2 forward.?

Now s = "zbd". Finally, shift the characters from index 0 to index 2 forward. Now s = "ace".

Example 2:

Input: s = "dztz", shifts = [[0,0,0],[1,1,1]]

Output: "catz"

Explanation: Firstly, shift the characters from index 0 to index 0 backward.?

Now s = "cztz". Finally, shift the characters from index 1 to index 1 forward.?

Now s = "catz".

?

Constraints:

• 1 <= s.length, shifts.length <= 5 * 104

• shifts[i].length == 3

• 0 <= starti?<= endi?< s.length

• 0 <= directioni?<= 1

• s?consists of lowercase English letters.

• 印象深刻的一道題啊，就是將更改from -to的位置記下來，向前則from++,to--(to--一直沒想明白，--就是抵消前面++的次數(shù))，然后一個變量累加，類似于前綴和，然后去遍歷，

• 中間出過一次問題，就是對數(shù)據(jù)%26之后還要加26再%26，就是避免溢出錯誤（還是別的負(fù)數(shù)的錯誤）我理解的是負(fù)數(shù)的錯誤。

Runtime:?10 ms, faster than?34.15%?of?Java?online submissions for?Shifting Letters II.

Memory Usage:?83.4 MB, less than?34.15%?of?Java?online submissions for?Shifting Letters II.