1
0
mirror of https://gitee.com/coder-xiaomo/leetcode-problemset synced 2025-09-14 03:41:41 +08:00
Code Issues Projects Releases Wiki Activity GitHub Gitee

add leetcode problem-cn part6

This commit is contained in:
2022-03-27 20:56:26 +08:00
parent c11d601602
commit 08d1f2a596
1096 changed files with 88216 additions and 2 deletions

View File

@@ -0,0 +1,33 @@
<p>编写一个高效的算法来判断 <code>m x n</code> 矩阵中,是否存在一个目标值。该矩阵具有如下特性:</p>
<ul>
<li>每行中的整数从左到右按升序排列。</li>
<li>每行的第一个整数大于前一行的最后一个整数。</li>
</ul>
<p> </p>
<p><strong>示例 1</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>输入:</strong>matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
<strong>输出:</strong>true
</pre>
<p><strong>示例 2</strong></p>
<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/11/25/mat2.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>输入:</strong>matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
<strong>输出:</strong>false
</pre>
<p> </p>
<p><strong>提示:</strong></p>
<ul>
<li><code>m == matrix.length</code></li>
<li><code>n == matrix[i].length</code></li>
<li><code>1 <= m, n <= 100</code></li>
<li><code>-10<sup>4</sup> <= matrix[i][j], target <= 10<sup>4</sup></code></li>
</ul>