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.
For 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.

在一个二维数组找一个元素，数组中的每一行和每一列都是升序排列的。我们可以从数组的右上方作者左下方的元素开始寻找。设定两个指针m, n，假设从右上方开始，如果matrix[m][n]与target相等就返回true；如果matrix[m][n]小于target就让m指针往下移动(m++); 如果大于target就让n指针向左移动(n--), 直到遍历完整个数组，这样时间复杂度为O(m+n)。代码如下：

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