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"translatedContent": "<p>你打算利用空闲时间来做兼职工作赚些零花钱。</p>\n\n<p>这里有&nbsp;<code>n</code>&nbsp;份兼职工作,每份工作预计从&nbsp;<code>startTime[i]</code>&nbsp;开始到&nbsp;<code>endTime[i]</code>&nbsp;结束,报酬为&nbsp;<code>profit[i]</code>。</p>\n\n<p>给你一份兼职工作表,包含开始时间&nbsp;<code>startTime</code>,结束时间&nbsp;<code>endTime</code>&nbsp;和预计报酬&nbsp;<code>profit</code>&nbsp;三个数组,请你计算并返回可以获得的最大报酬。</p>\n\n<p>注意,时间上出现重叠的 2 份工作不能同时进行。</p>\n\n<p>如果你选择的工作在时间&nbsp;<code>X</code>&nbsp;结束,那么你可以立刻进行在时间&nbsp;<code>X</code>&nbsp;开始的下一份工作。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/10/19/sample1_1584.png\" style=\"width: 300px;\"></strong></p>\n\n<pre><strong>输入:</strong>startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]\n<strong>输出:</strong>120\n<strong>解释:\n</strong>我们选出第 1 份和第 4 份工作, \n时间范围是 [1-3]+[3-6],共获得报酬 120 = 50 + 70。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/10/19/sample22_1584.png\" style=\"height: 112px; width: 600px;\"> </strong></p>\n\n<pre><strong>输入:</strong>startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]\n<strong>输出:</strong>150\n<strong>解释:\n</strong>我们选择第 145 份工作。 \n共获得报酬 150 = 20 + 70 + 60。\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<p><strong><img alt=\"\" src=\"https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2019/10/19/sample3_1584.png\" style=\"height: 112px; width: 400px;\"></strong></p>\n\n<pre><strong>输入:</strong>startTime = [1,1,1], endTime = [2,3,4], profit = [5,6,4]\n<strong>输出:</strong>6\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= startTime.length == endTime.length ==&nbsp;profit.length&nbsp;&lt;= 5 * 10^4</code></li>\n\t<li><code>1 &lt;=&nbsp;startTime[i] &lt;&nbsp;endTime[i] &lt;= 10^9</code></li>\n\t<li><code>1 &lt;=&nbsp;profit[i] &lt;= 10^4</code></li>\n</ul>\n",
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@@ -155,7 +155,7 @@
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"hints": [
"Think on DP.",
"Sort the elements by starting time, then define the dp[i] as the maximum profit taking elements from the suffix starting at i.",