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leetcode-cn/originData/sum-of-subarray-ranges.json
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"translatedContent": "<p>给你一个整数数组 <code>nums</code> 。<code>nums</code> 中，子数组的 <strong>范围</strong> 是子数组中最大元素和最小元素的差值。</p>\n\n<p>返回 <code>nums</code> 中 <strong>所有</strong> 子数组范围的 <strong>和</strong> <em>。</em></p>\n\n<p>子数组是数组中一个连续 <strong>非空</strong> 的元素序列。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,3]\n<strong>输出：</strong>4\n<strong>解释：</strong>nums 的 6 个子数组如下所示：\n[1]，范围 = 最大 - 最小 = 1 - 1 = 0 \n[2]，范围 = 2 - 2 = 0\n[3]，范围 = 3 - 3 = 0\n[1,2]，范围 = 2 - 1 = 1\n[2,3]，范围 = 3 - 2 = 1\n[1,2,3]，范围 = 3 - 1 = 2\n所有范围的和是 0 + 0 + 0 + 1 + 1 + 2 = 4</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,3,3]\n<strong>输出：</strong>4\n<strong>解释：</strong>nums 的 6 个子数组如下所示：\n[1]，范围 = 最大 - 最小 = 1 - 1 = 0\n[3]，范围 = 3 - 3 = 0\n[3]，范围 = 3 - 3 = 0\n[1,3]，范围 = 3 - 1 = 2\n[3,3]，范围 = 3 - 3 = 0\n[1,3,3]，范围 = 3 - 1 = 2\n所有范围的和是 0 + 0 + 0 + 2 + 0 + 2 = 4\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [4,-2,-3,4,1]\n<strong>输出：</strong>59\n<strong>解释：</strong>nums 中所有子数组范围的和是 59\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你可以设计一种时间复杂度为 <code>O(n)</code> 的解决方案吗？</p>\n",
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@@ -149,7 +149,7 @@
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"hints": [
"Can you get the max/min of a certain subarray by using the max/min of a smaller subarray within it?",
"Notice that the max of the subarray from index i to j is equal to max of (max of the subarray from index i to j-1) and nums[j]."