mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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183 lines
27 KiB
JSON
183 lines
27 KiB
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"title": "Find the K-or of an Array",
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"content": "<p>You are given an integer array <code>nums</code>, and an integer <code>k</code>. Let's introduce <strong>K-or</strong> operation by extending the standard bitwise OR. In K-or, a bit position in the result is set to <code>1</code> if at least <code>k</code> numbers in <code>nums</code> have a <code>1</code> in that position.</p>\n\n<p>Return <em>the K-or of</em> <code>nums</code>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1: </strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input:</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> nums = [7,12,9,8,9,15], k = 4 </span></p>\n\n<p><strong>Output:</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> 9 </span></p>\n\n<p><strong>Explanation: </strong></p>\n\n<p>Represent numbers in binary:</p>\n\n<table style=\"text-indent:10px; margin-bottom=20px;\">\n\t<tbody>\n\t\t<tr>\n\t\t\t<th><b>Number</b></th>\n\t\t\t<th>Bit 3</th>\n\t\t\t<th>Bit 2</th>\n\t\t\t<th>Bit 1</th>\n\t\t\t<th>Bit 0</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>7</b></td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>12</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>9</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>8</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>9</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>15</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>Result = 9</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<p>Bit 0 is set in 7, 9, 9, and 15. Bit 3 is set in 12, 9, 8, 9, and 15.<br />\nOnly bits 0 and 3 qualify. The result is <code>(1001)<sub>2</sub> = 9</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2: </strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input:</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> nums = [2,12,1,11,4,5], k = 6 </span></p>\n\n<p><strong>Output:</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> 0 </span></p>\n\n<p><strong>Explanation: </strong>No bit appears as 1 in all six array numbers, as required for K-or with <code>k = 6</code>. Thus, the result is 0.</p>\n</div>\n\n<p><strong class=\"example\">Example 3: </strong></p>\n\n<div class=\"example-block\" style=\"border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;\">\n<p><strong>Input:</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> nums = [10,8,5,9,11,6,8], k = 1 </span></p>\n\n<p><strong>Output:</strong><span class=\"example-io\" style=\"font-family: Menlo,sans-serif; font-size: 0.85rem;\"> 15 </span></p>\n\n<p><strong>Explanation: </strong> Since <code>k == 1</code>, the 1-or of the array is equal to the bitwise OR of all its elements. Hence, the answer is <code>10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15</code>.</p>\n</div>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 50</code></li>\n\t<li><code>0 <= nums[i] < 2<sup>31</sup></code></li>\n\t<li><code>1 <= k <= nums.length</code></li>\n</ul>\n",
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"translatedTitle": "找出数组中的 K-or 值",
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"translatedContent": "<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> 。让我们通过扩展标准的按位或来介绍 <strong>K-or</strong> 操作。在 K-or 操作中，如果在 <code>nums</code> 中，至少存在 <code>k</code> 个元素的第 <code>i</code> 位值为 1 ，那么 K-or 中的第 <code>i</code> 位的值是 1 。</p>\n\n<p>返回 <code>nums</code> 的 <strong>K-or</strong> 值。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [7,12,9,8,9,15], k = 4\n<strong>输出：</strong>9\n<strong>解释：</strong>\n用二进制表示 numbers：\n</pre>\n\n<table style=\"text-indent:10px; margin-bottom=20px;\">\n\t<tbody>\n\t\t<tr>\n\t\t\t<th><b>Number</b></th>\n\t\t\t<th>Bit 3</th>\n\t\t\t<th>Bit 2</th>\n\t\t\t<th>Bit 1</th>\n\t\t\t<th>Bit 0</th>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>7</b></td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>12</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>9</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>8</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>9</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>15</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t\t<tr>\n\t\t\t<td><b>Result = 9</b></td>\n\t\t\t<td>1</td>\n\t\t\t<td>0</td>\n\t\t\t<td>0</td>\n\t\t\t<td>1</td>\n\t\t</tr>\n\t</tbody>\n</table>\n\n<pre>\n位 0 在 7, 9, 9, 15 中为 1。位 3 在 12, 9, 8, 9, 15 中为 1。 只有位 0 和 3 满足。结果是 (1001)<sub>2</sub> = 9。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,12,1,11,4,5], k = 6\n<strong>输出：</strong>0\n<strong>解释：</strong>没有位在所有 6 个数字中都为 1，如 k = 6 所需要的。所以，答案为 0。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [10,8,5,9,11,6,8], k = 1\n<strong>输出：</strong>15\n<strong>解释：</strong>因为 k == 1 ，数组的 1-or 等于其中所有元素按位或运算的结果。因此，答案为 10 OR 8 OR 5 OR 9 OR 11 OR 6 OR 8 = 15 。</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 50</code></li>\n\t<li><code>0 <= nums[i] < 2<sup>31</sup></code></li>\n\t<li><code>1 <= k <= nums.length</code></li>\n</ul>\n",
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