mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
180 lines
19 KiB
JSON
180 lines
19 KiB
JSON
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"questionId": "1255",
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"questionFrontendId": "1330",
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"categoryTitle": "Algorithms",
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"boundTopicId": 79268,
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"title": "Reverse Subarray To Maximize Array Value",
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"content": "<p>You are given an integer array <code>nums</code>. The <em>value</em> of this array is defined as the sum of <code>|nums[i] - nums[i + 1]|</code> for all <code>0 <= i < nums.length - 1</code>.</p>\n\n<p>You are allowed to select any subarray of the given array and reverse it. You can perform this operation <strong>only once</strong>.</p>\n\n<p>Find maximum possible value of the final array.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,3,1,5,4]\n<strong>Output:</strong> 10\n<b>Explanation: </b>By reversing the subarray [3,1,5] the array becomes [2,5,1,3,4] whose value is 10.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,4,9,24,2,1,10]\n<strong>Output:</strong> 68\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code></li>\n</ul>\n",
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"translatedTitle": "翻转子数组得到最大的数组值",
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"translatedContent": "<p>给你一个整数数组 <code>nums</code> 。「数组值」定义为所有满足 <code>0 <= i < nums.length-1</code> 的 <code>|nums[i]-nums[i+1]|</code> 的和。</p>\n\n<p>你可以选择给定数组的任意子数组，并将该子数组翻转。但你只能执行这个操作 <strong>一次</strong> 。</p>\n\n<p>请你找到可行的最大 <strong>数组值 </strong>。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>nums = [2,3,1,5,4]\n<strong>输出：</strong>10\n<strong>解释：</strong>通过翻转子数组 [3,1,5] ，数组变成 [2,5,1,3,4] ，数组值为 10 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>nums = [2,4,9,24,2,1,10]\n<strong>输出：</strong>68\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 3*10^4</code></li>\n\t<li><code>-10^5 <= nums[i] <= 10^5</code></li>\n</ul>\n",
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"What's the score after reversing a sub-array [L, R] ?",
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"It's the score without reversing it + abs(a[R] - a[L-1]) + abs(a[L] - a[R+1]) - abs(a[L] - a[L-1]) - abs(a[R] - a[R+1])",
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"How to maximize that formula given that abs(x - y) = max(x - y, y - x) ?",
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"This can be written as max(max(a[R] - a[L - 1], a[L - 1] - a[R]) + max(a[R + 1] - a[L], a[L] - a[R + 1]) - value(L) - value(R + 1)) over all L < R where value(i) = abs(a[i] - a[i-1])",
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"This can be divided into 4 cases."
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