mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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190 lines
23 KiB
JSON
190 lines
23 KiB
JSON
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"categoryTitle": "Algorithms",
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"title": "Sum of Absolute Differences in a Sorted Array",
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"content": "<p>You are given an integer array <code>nums</code> sorted in <strong>non-decreasing</strong> order.</p>\n\n<p>Build and return <em>an integer array </em><code>result</code><em> with the same length as </em><code>nums</code><em> such that </em><code>result[i]</code><em> is equal to the <strong>summation of absolute differences</strong> between </em><code>nums[i]</code><em> and all the other elements in the array.</em></p>\n\n<p>In other words, <code>result[i]</code> is equal to <code>sum(|nums[i]-nums[j]|)</code> where <code>0 <= j < nums.length</code> and <code>j != i</code> (<strong>0-indexed</strong>).</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,3,5]\n<strong>Output:</strong> [4,3,5]\n<strong>Explanation:</strong> Assuming the arrays are 0-indexed, then\nresult[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,\nresult[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,\nresult[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,4,6,8,10]\n<strong>Output:</strong> [24,15,13,15,21]\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= nums[i + 1] <= 10<sup>4</sup></code></li>\n</ul>\n",
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"translatedTitle": "有序数组中差绝对值之和",
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"translatedContent": "<p>给你一个 <strong>非递减 </strong>有序整数数组 <code>nums</code> 。</p>\n\n<p>请你建立并返回一个整数数组<em> </em><code>result</code>，它跟<em> </em><code>nums</code> 长度相同，且<code>result[i]</code> 等于<em> </em><code>nums[i]</code> 与数组中所有其他元素差的绝对值之和。</p>\n\n<p>换句话说， <code>result[i]</code> 等于 <code>sum(|nums[i]-nums[j]|)</code> ，其中 <code>0 <= j < nums.length</code> 且 <code>j != i</code> （下标从 0 开始）。</p>\n\n<p> </p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [2,3,5]\n<b>输出：</b>[4,3,5]\n<b>解释：</b>假设数组下标从 0 开始，那么\nresult[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4，\nresult[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3，\nresult[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>nums = [1,4,6,8,10]\n<b>输出：</b>[24,15,13,15,21]\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= nums[i] <= nums[i + 1] <= 10<sup>4</sup></code></li>\n</ul>\n",
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"Absolute difference is the same as max(a, b) - min(a, b). How can you use this fact with the fact that the array is sorted?",
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"For nums[i], the answer is (nums[i] - nums[0]) + (nums[i] - nums[1]) + ... + (nums[i] - nums[i-1]) + (nums[i+1] - nums[i]) + (nums[i+2] - nums[i]) + ... + (nums[n-1] - nums[i]).",
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"It can be simplified to (nums[i] * i - (nums[0] + nums[1] + ... + nums[i-1])) + ((nums[i+1] + nums[i+2] + ... + nums[n-1]) - nums[i] * (n-i-1)). One can build prefix and suffix sums to compute this quickly."
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