search-a-2d-matrix

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Algorithms

You are given an <code>m x n</code> integer matrix <code>matrix</code> with the following two properties:

<ul>
	<li>Each row is sorted in non-decreasing order.</li>
	<li>The first integer of each row is greater than the last integer of the previous row.</li>
</ul>

Given an integer <code>target</code>, return <code>true</code> if <code>target</code> is in <code>matrix</code> or <code>false</code> otherwise.

You must write a solution in <code>O(log(m * n))</code> time complexity.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat.jpg" style="width: 322px; height: 242px;" />
<pre>
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
Output: true
</pre>

Example 2:
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat2.jpg" style="width: 322px; height: 242px;" />
<pre>
Input: matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
Output: false
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>-104 &lt;= matrix[i][j], target &lt;= 104</code></li>
</ul>

searchMatrix

Search a 2D Matrix

Search a 2D Matrix II

search-a-2d-matrix-ii

搜索二维矩阵 II

Split Message Based on Limit

split-message-based-on-limit

根据限制分割消息

Array

Binary Search

Matrix

给你一个满足下述两条属性的 <code>m x n</code> 整数矩阵：

<ul>
	<li>每行中的整数从左到右按非严格递增顺序排列。</li>
	<li>每行的第一个整数大于前一行的最后一个整数。</li>
</ul>

给你一个整数 <code>target</code> ，如果 <code>target</code> 在矩阵中，返回 <code>true</code> ；否则，返回 <code>false</code> 。

&nbsp;

示例 1：
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/05/mat.jpg" style="width: 322px; height: 242px;" />
<pre>
输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
输出：true
</pre>

示例 2：
<img alt="" src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2020/11/25/mat2.jpg" style="width: 322px; height: 242px;" />
<pre>
输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
输出：false
</pre>

&nbsp;

提示：

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>-104 &lt;= matrix[i][j], target &lt;= 104</code></li>
</ul>