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.

思路：

这个技巧很明显，先对比最后一个，找到是在哪行，然后再遍历这个行来对比（行内可以用binary search）

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