mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
41 lines
1.7 KiB
HTML
41 lines
1.7 KiB
HTML
<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>
|
|
|
|
<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>
|
|
|
|
<p>Return<em> an array containing the index of the row, and the number of ones in it.</em></p>
|
|
|
|
<p> </p>
|
|
<p><strong class="example">Example 1:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> mat = [[0,1],[1,0]]
|
|
<strong>Output:</strong> [0,1]
|
|
<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].
|
|
</pre>
|
|
|
|
<p><strong class="example">Example 2:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> mat = [[0,0,0],[0,1,1]]
|
|
<strong>Output:</strong> [1,2]
|
|
<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].
|
|
</pre>
|
|
|
|
<p><strong class="example">Example 3:</strong></p>
|
|
|
|
<pre>
|
|
<strong>Input:</strong> mat = [[0,0],[1,1],[0,0]]
|
|
<strong>Output:</strong> [1,2]
|
|
<strong>Explanation:</strong> The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].
|
|
</pre>
|
|
|
|
<p> </p>
|
|
<p><strong>Constraints:</strong></p>
|
|
|
|
<ul>
|
|
<li><code>m == mat.length</code> </li>
|
|
<li><code>n == mat[i].length</code> </li>
|
|
<li><code>1 <= m, n <= 100</code> </li>
|
|
<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
|
|
</ul>
|