mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-09-04 06:51:41 +08:00
175 lines
22 KiB
JSON
175 lines
22 KiB
JSON
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"questionId": "1000476",
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"questionFrontendId": "LCP 61",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1845919,
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"title": "气温变化趋势",
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"titleSlug": "6CE719",
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"content": "English description is not available for the problem. Please switch to Chinese.",
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"translatedTitle": "气温变化趋势",
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"translatedContent": "力扣城计划在两地设立「力扣嘉年华」的分会场,气象小组正在分析两地区的气温变化趋势,对于第 `i ~ (i+1)` 天的气温变化趋势,将根据以下规则判断:\n- 若第 `i+1` 天的气温 **高于** 第 `i` 天,为 **上升** 趋势\n- 若第 `i+1` 天的气温 **等于** 第 `i` 天,为 **平稳** 趋势\n- 若第 `i+1` 天的气温 **低于** 第 `i` 天,为 **下降** 趋势\n\n已知 `temperatureA[i]` 和 `temperatureB[i]` 分别表示第 `i` 天两地区的气温。\n组委会希望找到一段天数尽可能多,且两地气温变化趋势相同的时间举办嘉年华活动。请分析并返回两地气温变化趋势**相同的最大连续天数**。\n> 即最大的 `n`,使得第 `i~i+n` 天之间,两地气温变化趋势相同\n\n**示例 1:**\n>输入:\n>`temperatureA = [21,18,18,18,31]`\n>`temperatureB = [34,32,16,16,17]`\n>\n>输出:`2`\n>\n>解释:如下表所示, 第 `2~4` 天两地气温变化趋势相同,且持续时间最长,因此返回 `4-2=2`\n{:width=1000px}\n\n\n**示例 2:**\n>输入:\n>`temperatureA = [5,10,16,-6,15,11,3]`\n>`temperatureB = [16,22,23,23,25,3,-16]`\n>\n>输出:`3`\n\n**提示:**\n- `2 <= temperatureA.length == temperatureB.length <= 1000`\n- `-20 <= temperatureA[i], temperatureB[i] <= 40`\n",
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"difficulty": "Easy",
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"likes": 35,
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"username": "LeetCode",
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"code": "class Solution {\npublic:\n int temperatureTrend(vector<int>& temperatureA, vector<int>& temperatureB) {\n\n }\n};",
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"code": "class Solution(object):\n def temperatureTrend(self, temperatureA, temperatureB):\n \"\"\"\n :type temperatureA: List[int]\n :type temperatureB: List[int]\n :rtype: int\n \"\"\"",
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"code": "class Solution {\n\n /**\n * @param Integer[] $temperatureA\n * @param Integer[] $temperatureB\n * @return Integer\n */\n function temperatureTrend($temperatureA, $temperatureB) {\n\n }\n}",
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"lang": "Kotlin",
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"code": "class Solution {\n fun temperatureTrend(temperatureA: IntArray, temperatureB: IntArray): Int {\n\n }\n}",
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"code": "# @param {Integer[]} temperature_a\n# @param {Integer[]} temperature_b\n# @return {Integer}\ndef temperature_trend(temperature_a, temperature_b)\n\nend",
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