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.

觀察到，左下角的數字，向上遞減，向右遞增，可以從左下角開始查詢，迴圈成立條件是陣列下標不越界。如果 當前數 < target，向右查詢；如果 當前數 < target，向上查詢。

time: O(M+N), space: O(1)

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
        int row = matrix.length - 1, col = 0;
        while(row >= 0 && col <= matrix[0].length - 1) {
            
 
if(matrix[row][col] == target) return true;
            if(matrix[row][col] > target) row--;
            else col++;
        }
        return false;
    }
}