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leetcode-problemset/leetcode-cn/originData/t3fKg1.json
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{
"data": {
"question": {
"questionId": "1000219",
"questionFrontendId": "LCP 32",
"categoryTitle": "Algorithms",
"boundTopicId": 701752,
"title": "批量处理任务",
"titleSlug": "t3fKg1",
"content": "English description is not available for the problem. Please switch to Chinese.",
"translatedTitle": "批量处理任务",
"translatedContent": "某实验室计算机待处理任务以 `[start,end,period]` 格式记于二维数组 `tasks`,表示完成该任务的时间范围为起始时间 `start` 至结束时间 `end` 之间,需要计算机投入 `period` 的时长,注意:\n1. `period` 可为不连续时间\n2. 首尾时间均包含在内\n\n处于开机状态的计算机可同时处理任意多个任务请返回电脑最少开机多久可处理完所有任务。\n\n**示例 1**\n>输入:`tasks = [[1,3,2],[2,5,3],[5,6,2]]`\n>\n>输出:`4`\n>\n>解释:\n>tasks[0] 选择时间点 2、3\n>tasks[1] 选择时间点 2、3、5\n>tasks[2] 选择时间点 5、6\n>因此计算机仅需在时间点 2、3、5、6 四个时刻保持开机即可完成任务。\n\n**示例 2**\n>输入:`tasks = [[2,3,1],[5,5,1],[5,6,2]]`\n>\n>输出:`3`\n>\n>解释:\n>tasks[0] 选择时间点 2 或 3\n>tasks[1] 选择时间点 5\n>tasks[2] 选择时间点 5、6\n>因此计算机仅需在时间点 2、5、6 或 3、5、6 三个时刻保持开机即可完成任务。\n\n**提示:**\n- `2 <= tasks.length <= 10^5`\n- `tasks[i].length == 3`\n- `0 <= tasks[i][0] <= tasks[i][1] <= 10^9`\n- `1 <= tasks[i][2] <= tasks[i][1]-tasks[i][0] + 1`",
"isPaidOnly": false,
"difficulty": "Hard",
"likes": 43,
"dislikes": 0,
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"similarQuestions": "[]",
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"name": "Greedy",
"slug": "greedy",
"translatedName": "贪心",
"__typename": "TopicTagNode"
},
{
"name": "Array",
"slug": "array",
"translatedName": "数组",
"__typename": "TopicTagNode"
},
{
"name": "Heap (Priority Queue)",
"slug": "heap-priority-queue",
"translatedName": "堆(优先队列)",
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"code": "class Solution {\npublic:\n int processTasks(vector<vector<int>>& tasks) {\n\n }\n};",
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"code": "class Solution {\n public int processTasks(int[][] tasks) {\n\n }\n}",
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"code": "class Solution(object):\n def processTasks(self, tasks):\n \"\"\"\n :type tasks: List[List[int]]\n :rtype: int\n \"\"\"",
"__typename": "CodeSnippetNode"
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"code": "class Solution:\n def processTasks(self, tasks: List[List[int]]) -> int:",
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"code": "\n\nint processTasks(int** tasks, int tasksSize, int* tasksColSize){\n\n}",
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