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.

题意：三个条件，

1. 数字应该从左到右的增加
2. 每行的第一个也应该是个递增的
3. 给一个target看这里面是否包含

都满足就可以了。

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        boolean isHaveTarget = false;
        for (int i = 0; i < matrix.length - 1; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                if (matrix[i][j] >= matrix[i + 1][j]) {
                    return false;
                }
            }
        }

        for (int i = 0; i < matrix.length - 1; i++) {
            if (matrix[i][0] > matrix[i + 1][0]) {    
                return false;
            }
        }

        for (int i = 0; i < matrix.length; i++) {
            for (int j = 0; j < matrix[0].length; j++) {
                if (matrix[i][j] == target) {
                    isHaveTarget = true;
                }
            }
        return isHaveTarget;
    }
}