74. Search a 2D Matrix

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

Integers in each row are sorted from left to right.

The first integer of each row is greater than the last integer of the previous row.

For example,

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

在一个m*n二维数组中，每一行从左到右递增，每一行的第一个元素比上一行最后一个元素大。

判断某个元素是否在该数组中。

 1     public boolean searchMatrix(int[][] matrix, int target) {
 2         if (matrix.length == 0 || matrix[0].length == 0) return false;
 3         if (target < matrix[0][0] || target > matrix[matrix.length - 1][matrix[0].length - 1]) return false;
 4         int m = matrix.length;
 5         int n = matrix[0].length;
 6         int left = 0, right = m * n - 1;
 7         while (left < right) {
 8             int mid = left + (right - left) / 2;
 9             if (matrix[mid / n][mid % n] == target) return true;
10             else if (matrix[mid / n][mid % n] > target) right = mid - 1;
11             else left = mid + 1;
12         }
13         return matrix[left / n][left % n] == target;
14     }