{ "data": { "question": { "questionId": "3164", "questionFrontendId": "2899", "categoryTitle": "Algorithms", "boundTopicId": 2480185, "title": "Last Visited Integers", "titleSlug": "last-visited-integers", "content": "
Given a 0-indexed array of strings words
where words[i]
is either a positive integer represented as a string or the string "prev"
.
Start iterating from the beginning of the array; for every "prev"
string seen in words
, find the last visited integer in words
which is defined as follows:
k
be the number of consecutive "prev"
strings seen so far (containing the current string). Let nums
be the 0-indexed array of integers seen so far and nums_reverse
be the reverse of nums
, then the integer at (k - 1)th
index of nums_reverse
will be the last visited integer for this "prev"
.k
is greater than the total visited integers, then the last visited integer will be -1
.Return an integer array containing the last visited integers.
\n\n\n
Example 1:
\n\n\nInput: words = ["1","2","prev","prev","prev"]\nOutput: [2,1,-1]\nExplanation: \nFor "prev" at index = 2, last visited integer will be 2 as here the number of consecutive "prev" strings is 1, and in the array reverse_nums, 2 will be the first element.\nFor "prev" at index = 3, last visited integer will be 1 as there are a total of two consecutive "prev" strings including this "prev" which are visited, and 1 is the second last visited integer.\nFor "prev" at index = 4, last visited integer will be -1 as there are a total of three consecutive "prev" strings including this "prev" which are visited, but the total number of integers visited is two.\n\n\n
Example 2:
\n\n\nInput: words = ["1","prev","2","prev","prev"]\nOutput: [1,2,1]\nExplanation:\nFor "prev" at index = 1, last visited integer will be 1.\nFor "prev" at index = 3, last visited integer will be 2.\nFor "prev" at index = 4, last visited integer will be 1 as there are a total of two consecutive "prev" strings including this "prev" which are visited, and 1 is the second last visited integer.\n\n\n
\n
Constraints:
\n\n1 <= words.length <= 100
words[i] == "prev"
or 1 <= int(words[i]) <= 100
给你一个下标从 0 开始的字符串数组 words
,其中 words[i]
要么是一个字符串形式的正整数,要么是字符串 \"prev\"
。
我们从数组的开头开始遍历,对于 words
中的每个 \"prev\"
字符串,找到 words
中的 上一个遍历的整数 ,定义如下:
k
表示到当前位置为止的连续 \"prev\"
字符串数目(包含当前字符串),令下标从 0 开始的 整数 数组 nums
表示目前为止遍历过的所有整数,同时用 nums_reverse
表示 nums
反转得到的数组,那么当前 \"prev\"
对应的 上一个遍历的整数 是 nums_reverse
数组中下标为 (k - 1)
的整数。k
比目前为止遍历过的整数数目 更多 ,那么上一个遍历的整数为 -1
。请你返回一个整数数组,包含所有上一个遍历的整数。
\n\n\n\n
示例 1:
\n\n\n输入:words
= [\"1\",\"2\",\"prev\",\"prev\",\"prev\"]\n输出:[2,1,-1]\n解释:\n对于下标为 2 处的 \"prev\" ,上一个遍历的整数是 2 ,因为连续 \"prev\" 数目为 1 ,同时在数组 reverse_nums 中,第一个元素是 2 。\n对于下标为 3 处的 \"prev\" ,上一个遍历的整数是 1 ,因为连续 \"prev\" 数目为 2 ,同时在数组 reverse_nums 中,第二个元素是 1 。\n对于下标为 4 处的 \"prev\" ,上一个遍历的整数是 -1 ,因为连续 \"prev\" 数目为 3 ,但总共只遍历过 2 个整数。\n
\n\n示例 2:
\n\n\n输入:words
= [\"1\",\"prev\",\"2\",\"prev\",\"prev\"]\n输出:[1,2,1]\n解释:\n对于下标为 1 处的 \"prev\" ,上一个遍历的整数是 1 。\n对于下标为 3 处的 \"prev\" ,上一个遍历的整数是 2 。\n对于下标为 4 处的 \"prev\" ,上一个遍历的整数是 1 ,因为连续 \"prev\" 数目为 2 ,同时在数组 reverse_nums 中,第二个元素是 1 。\n
\n\n\n\n
提示:
\n\n1 <= words.length <= 100
words[i] == \"prev\"
或 1 <= int(words[i]) <= 100
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