74:搜索二维矩阵

src/main/java/leetcode/editor/cn/SearchA2dMatrix.java

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+//编写一个高效的算法来判断 m x n 矩阵中，是否存在一个目标值。该矩阵具有如下特性： 
+//
+// 
+// 每行中的整数从左到右按升序排列。 
+// 每行的第一个整数大于前一行的最后一个整数。 
+// 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 3
+//输出：true
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：matrix = [[1,3,5,7],[10,11,16,20],[23,30,34,60]], target = 13
+//输出：false
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// m == matrix.length 
+// n == matrix[i].length 
+// 1 <= m, n <= 100 
+// -104 <= matrix[i][j], target <= 104 
+// 
+// Related Topics 数组 二分查找 矩阵 
+// 👍 478 👎 0
+
+package leetcode.editor.cn;
+
+import com.code.leet.entiy.TwoArray;
+
+//74:搜索二维矩阵
+class SearchA2dMatrix {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new SearchA2dMatrix().new Solution();
+        //true
+//        TwoArray twoArray = new TwoArray("[[1,3,5,7],[10,11,16,20],[23,30,34,60]]",true);
+//        System.out.println(solution.searchMatrix(twoArray.getArr(), 3));
+        //true
+        TwoArray twoArray = new TwoArray("[[1,3]]", true);
+        System.out.println(solution.searchMatrix(twoArray.getArr(), 3));
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public boolean searchMatrix(int[][] matrix, int target) {
+            if (target < matrix[0][0] || target > matrix[matrix.length - 1][matrix[0].length - 1]) {
+                return false;
+            }
+            int start = 0;
+            int end = matrix.length - 1;
+            while (start < end - 1) {
+                int mid = (start + end) / 2;
+                if (matrix[mid][0] == target) {
+                    return true;
+                }
+                if (matrix[mid][0] > target) {
+                    end = mid - 1;
+                } else {
+                    start = mid;
+                }
+            }
+            if (matrix[end][0] == target) {
+                return true;
+            }
+            int xIndex = matrix[end][0] > target ? start : end;
+            start = 0;
+            end = matrix[0].length - 1;
+            while (start < end) {
+                int mid = (start + end) / 2;
+                if (matrix[xIndex][mid] == target) {
+                    return true;
+                }
+                if (matrix[xIndex][mid] < target) {
+                    start = mid + 1;
+                } else {
+                    end = mid;
+                }
+            }
+            return matrix[xIndex][start] == target;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}

src/main/java/leetcode/editor/cn/SearchA2dMatrix.md

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+<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>
+<div><div>Related Topics</div><div><li>数组</li><li>二分查找</li><li>矩阵</li></div></div>\n<div><li>👍 478</li><li>👎 0</li></div>