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‎README.adoc

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@@ -544,12 +544,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
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|Medium
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//|{counter:codes}
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//|{leetcode_base_url}/search-a-2d-matrix/[74. Search a 2D Matrix^]
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//|{source_base_url}/_0074_SearchA2DMatrix.java[Java]
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//|{doc_base_url}/0074-search-a-2d-matrix.adoc[题解]
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//|Medium
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//|
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|{counter:codes}
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|{leetcode_base_url}/search-a-2d-matrix/[74. Search a 2D Matrix^]
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|{source_base_url}/_0074_SearchA2DMatrix.java[Java]
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|{doc_base_url}/0074-search-a-2d-matrix.adoc[题解]
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|Medium
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|
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|{counter:codes}
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|{leetcode_base_url}/sort-colors/[75. Sort Colors^]

‎docs/0074-search-a-2d-matrix.adoc

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[#0074-search-a-2d-matrix]
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= 74. Search a 2D Matrix
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= 74. 搜索二维矩阵
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{leetcode}/problems/search-a-2d-matrix/[LeetCode - Search a 2D Matrix^]
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https://leetcode.cn/problems/search-a-2d-matrix/[LeetCode - 74. 搜索二维矩阵 ^]
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Write an efficient algorithm that searches for a value in an _m_ x _n_ matrix. This matrix has the following properties:
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给你一个满足下述两条属性的 `m x n` 整数矩阵：
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* 每行中的整数从左到右按非严格递增顺序排列。
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* 每行的第一个整数大于前一行的最后一个整数。
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* Integers in each row are sorted from left to right.
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* The first integer of each row is greater than the last integer of the previous row.
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给你一个整数 `target` ，如果 `target` 在矩阵中，返回 `true`；否则，返回 `false` 。
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*示例 1：*
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*Example 1:*
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image::images/0074-01.jpg[{image_attr}]
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[subs="verbatim,quotes,macros"]
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----
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*Input:*
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matrix = [
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[1, 3, 5, 7],
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[10, 11, 16, 20],
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[23, 30, 34, 50]
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]
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target = 3
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*Output:* true
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....
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输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
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输出：true
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....
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----
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*示例 2：*
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*Example 2:*
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image::images/0074-02.jpg[{image_attr}]
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[subs="verbatim,quotes,macros"]
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----
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*Input:*
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matrix = [
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[1, 3, 5, 7],
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[10, 11, 16, 20],
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[23, 30, 34, 50]
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]
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target = 13
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*Output:* false
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----
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....
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输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
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输出：false
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....
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*提示：*
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* `m == matrix.length`
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* `n == matrix[i].length`
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* `+1 <= m, n <= 100+`
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* `-10^4^ \<= matrix[i][j], target \<= 10^4^`
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== 思路分析
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由于矩阵同行有序，下一行比上一行大，所以，把每一行“拼接”起来就是一个有序数组，可以用二分查找解决问题。
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[[src-0074]]
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[tabs]
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====
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一刷::
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--
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[{java_src_attr}]
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----
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include::{sourcedir}/_0074_SearchA2DMatrix.java[tag=answer]
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----
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--
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// 二刷::
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// +
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// --
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// [{java_src_attr}]
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// ----
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// include::{sourcedir}/_0074_SearchA2DMatrix_2.java[tag=answer]
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// ----
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// --
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====
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== 参考资料
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. https://leetcode.cn/problems/search-a-2d-matrix/solutions/688117/sou-suo-er-wei-ju-zhen-by-leetcode-solut-vxui/[74. 搜索二维矩阵 - 官方题解^]

‎docs/images/0074-01.jpg

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‎docs/images/0074-02.jpg

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‎docs/index.adoc

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// include::0074-search-a-2d-matrix.adoc[leveloffset=+1]
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include::0074-search-a-2d-matrix.adoc[leveloffset=+1]
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‎logbook/202503.adoc

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|{doc_base_url}/0032-longest-valid-parentheses.adoc[题解]
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|⭕️ 栈。使用栈记录左右括号的下标，匹配后标注“占位符”。最后，统计占位符中的数量，找出最大长度。
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|{counter:codes2503}
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|{leetcode_base_url}/search-a-2d-matrix/[74. 搜索二维矩阵^]
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|{doc_base_url}/0074-search-a-2d-matrix.adoc[题解]
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|✅ 二分查找。把矩阵按行“拼接”，然后二分查找，解法非常妙。
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|===
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截止目前，本轮练习一共完成 {codes2503} 道题。

‎src/main/java/com/diguage/algo/leetcode/_0073_SetMatrixZeroes.java

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* Runtime: 1 ms, faster than 100.00% of Java online submissions for Set Matrix Zeroes.
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*
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* Memory Usage: 50.4 MB, less than 5.71% of Java online submissions for Set Matrix Zeroes.
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*
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* @author D瓜哥 · https://www.diguage.com
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* @since 2019-10-27 00:59
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*/
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public void setZeroes(int[][] matrix) {
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if (Objects.isNull(matrix) || matrix[0].length == 0) {

‎src/main/java/com/diguage/algo/leetcode/_0074_SearchA2DMatrix.java

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package com.diguage.algo.leetcode;
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public class _0074_SearchA2DMatrix {
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// tag::answer[]
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/**
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* @author D瓜哥 · https://www.diguage.com
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* @since 2025-04-23 18:20:50
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*/
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public boolean searchMatrix(int[][] matrix, int target) {
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int row = matrix.length;
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int column = matrix[0].length;
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int left = 0, right = row * column - 1;
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while (left <= right) {
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int mid = left + (right - left) / 2;
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int num = matrix[mid / column][mid % column];
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if (num == target) {
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return true;
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} else if (target < num) {
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right = mid - 1;
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} else {
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left = mid + 1;
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}
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}
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return false;
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}
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// end::answer[]
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public static void main(String[] args) {
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new _0074_SearchA2DMatrix().searchMatrix(new int[][]{
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{1, 3, 5, 7},
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{10, 11, 16, 20},
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{23, 30, 34, 60}},
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3);
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}
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}

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