{ "data": { "question": { "questionId": "2465", "questionFrontendId": "2381", "categoryTitle": "Algorithms", "boundTopicId": 1759858, "title": "Shifting Letters II", "titleSlug": "shifting-letters-ii", "content": "

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.

\n\n

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').

\n\n

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

\n\n

 

\n

Example 1:

\n\n
\nInput: s = "abc", shifts = [[0,1,0],[1,2,1],[0,2,1]]\nOutput: "ace"\nExplanation: Firstly, shift the characters from index 0 to index 1 backward. Now s = "zac".\nSecondly, shift the characters from index 1 to index 2 forward. Now s = "zbd".\nFinally, shift the characters from index 0 to index 2 forward. Now s = "ace".
\n\n

Example 2:

\n\n
\nInput: s = "dztz", shifts = [[0,0,0],[1,1,1]]\nOutput: "catz"\nExplanation: Firstly, shift the characters from index 0 to index 0 backward. Now s = "cztz".\nFinally, shift the characters from index 1 to index 1 forward. Now s = "catz".\n
\n\n

 

\n

Constraints:

\n\n\n", "translatedTitle": "字母移位 II", "translatedContent": "

给你一个小写英文字母组成的字符串 s 和一个二维整数数组 shifts ,其中 shifts[i] = [starti, endi, directioni] 。对于每个 i ,将 s 中从下标 starti 到下标 endi (两者都包含)所有字符都进行移位运算,如果 directioni = 1 将字符向后移位,如果 directioni = 0 将字符向前移位。

\n\n

将一个字符 向后 移位的意思是将这个字符用字母表中 下一个 字母替换(字母表视为环绕的,所以 'z' 变成 'a')。类似的,将一个字符 向前 移位的意思是将这个字符用字母表中 前一个 字母替换(字母表是环绕的,所以 'a' 变成 'z' )。

\n\n

请你返回对 s 进行所有移位操作以后得到的最终字符串。

\n\n

 

\n\n

示例 1:

\n\n
输入:s = \"abc\", shifts = [[0,1,0],[1,2,1],[0,2,1]]\n输出:\"ace\"\n解释:首先,将下标从 0 到 1 的字母向前移位,得到 s = \"zac\" 。\n然后,将下标从 1 到 2 的字母向后移位,得到 s = \"zbd\" 。\n最后,将下标从 0 到 2 的字符向后移位,得到 s = \"ace\" 。
\n\n

示例 2:

\n\n
输入:s = \"dztz\", shifts = [[0,0,0],[1,1,1]]\n输出:\"catz\"\n解释:首先,将下标从 0 到 0 的字母向前移位,得到 s = \"cztz\" 。\n最后,将下标从 1 到 1 的字符向后移位,得到 s = \"catz\" 。\n
\n\n

 

\n\n

提示:

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\\u5df2\\u9884\\u5148 (require data\\/gvector data\\/queue data\\/order data\\/heap). \\u82e5\\u9700\\u4f7f\\u7528\\u5176\\u5b83\\u6570\\u636e\\u7ed3\\u6784\\uff0c\\u53ef\\u81ea\\u884c require\\u3002<\\/p>\"],\"erlang\":[\"Erlang\",\"Erlang\\/OTP 26\"],\"elixir\":[\"Elixir\",\"Elixir 1.17 with Erlang\\/OTP 26\"],\"dart\":[\"Dart\",\"

Dart 3.2\\u3002\\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 collection<\\/a> \\u5305<\\/p>\\r\\n\\r\\n

\\u60a8\\u7684\\u4ee3\\u7801\\u5c06\\u4f1a\\u88ab\\u4e0d\\u7f16\\u8bd1\\u76f4\\u63a5\\u8fd0\\u884c<\\/p>\"],\"cangjie\":[\"Cangjie\",\"

\\u7248\\u672c\\uff1a0.53.11 (cjnative)<\\/p>\\r\\n\\r\\n

\\u7f16\\u8bd1\\u53c2\\u6570\\uff1a-O2 --disable-reflection<\\/code><\\/p>\\r\\n\\r\\n

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