240. Search a 2D Matrix II

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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
Example:

Consider the following matrix:
[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]
Given target = 5, return true.
Given target = 20, return false

难度：medium

题目：设计高效算法在矩阵中搜索给定的值。矩阵特性如下：每行元素从左到右递增，每列元素从上到下递增。

思路：在第一个行中找出首元素大于target的列，在第一个列中找出首元素大于target的行。然后对每行进行二叉搜索。

Runtime: 6 ms, faster than 97.21% of Java online submissions for Search a 2D Matrix II.
Memory Usage: 46.6 MB, less than 23.14% of Java online submissions for Search a 2D Matrix II.

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length == 0 || matrix[0].length == 0) {
            return false;
        }
        
        int m = matrix.length, n = matrix[0].length;
        int row = 0, column = 0;
        for (; column < n; column++) {
            if (matrix[0][column] > target) {
                break;
            }
        }
        for (; row < m; row++) {
            if (matrix[row][0] > target) {
                break;
            }
        }
        
        for (int i = 0; i < row; i++) {
            if (Arrays.binarySearch(matrix[i], 0, column, target) >= 0) {
                return true;
            }
        }
        return false;
    }
}