mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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176 lines
22 KiB
JSON
176 lines
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"title": "Row With Maximum Ones",
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"content": "<p>Given a <code>m x n</code> binary matrix <code>mat</code>, find the <strong>0-indexed</strong> position of the row that contains the <strong>maximum</strong> count of <strong>ones,</strong> and the number of ones in that row.</p>\n\n<p>In case there are multiple rows that have the maximum count of ones, the row with the <strong>smallest row number</strong> should be selected.</p>\n\n<p>Return<em> an array containing the index of the row, and the number of ones in it.</em></p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> mat = [[0,1],[1,0]]\n<strong>Output:</strong> [0,1]\n<strong>Explanation:</strong> Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1<code>)</code>. So, the answer is [0,1]. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> mat = [[0,0,0],[0,1,1]]\n<strong>Output:</strong> [1,2]\n<strong>Explanation:</strong> The row indexed 1 has the maximum count of ones <code>(2)</code>. So we return its index, <code>1</code>, and the count. So, the answer is [1,2].\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> mat = [[0,0],[1,1],[0,0]]\n<strong>Output:</strong> [1,2]\n<strong>Explanation:</strong> The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == mat.length</code> </li>\n\t<li><code>n == mat[i].length</code> </li>\n\t<li><code>1 <= m, n <= 100</code> </li>\n\t<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>\n</ul>\n",
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"translatedTitle": "一最多的行",
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"translatedContent": "<p>给你一个大小为 <code>m x n</code> 的二进制矩阵 <code>mat</code> ,请你找出包含最多 <strong>1</strong> 的行的下标(从 <strong>0</strong> 开始)以及这一行中 <strong>1</strong> 的数目。</p>\n\n<p>如果有多行包含最多的 1 ,只需要选择 <strong>行下标最小</strong> 的那一行。</p>\n\n<p>返回一个由行下标和该行中 1 的数量组成的数组。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>mat = [[0,1],[1,0]]\n<strong>输出:</strong>[0,1]\n<strong>解释:</strong>两行中 1 的数量相同。所以返回下标最小的行,下标为 0 。该行 1 的数量为 1 。所以,答案为 [0,1] 。 \n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>mat = [[0,0,0],[0,1,1]]\n<strong>输出:</strong>[1,2]\n<strong>解释:</strong>下标为 1 的行中 1 的数量最多<code>。</code>该行 1 的数量<code>为 2 。所以,答案为</code> [1,2] 。\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>mat = [[0,0],[1,1],[0,0]]\n<strong>输出:</strong>[1,2]\n<strong>解释:</strong>下标为 1 的行中 1 的数量最多。该行 1 的数量<code>为 2 。所以,答案为</code> [1,2] 。</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>m == mat.length</code> </li>\n\t<li><code>n == mat[i].length</code> </li>\n\t<li><code>1 <= m, n <= 100</code> </li>\n\t<li><code>mat[i][j]</code> 为 <code>0</code> 或 <code>1</code></li>\n</ul>\n",
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