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.

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.

题目含义：矩阵的行和列都是升序排列，查找一个数是否在矩阵中

 1     public boolean searchMatrix(int[][] matrix, int target) {
 2 //        从右上角开始, 比较target 和 matrix[i][j]的值. 先列后行来逼近。如果大于target, 则该列不可能有此数, 所以j--。如果小于target, 则该行不可能有此数,  所以i++;
 4         if (matrix.length == 0 || matrix[0].length == 0) return false;
 5         int i = 0, j = matrix[0].length - 1;
 6         while (i < matrix.length && j >= 0) {
 7             int x = matrix[i][j];
 8             if (target == x) return true;
 9             else if (target < x) j--;
10             else i++;
11         }
12         return false;     
13     }