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.

class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix.length == 0 || matrix[0].length == 0)
            return false;
        int targetRow = 0;
        //設定臨時陣列儲存每行第一列數字
        int[] temp = new int[matrix.length];
        for(int i = 0;i < matrix.length;i ++)
            temp[i] = matrix[i][0];
        //找到目標元素所在行
        targetRow = binarySearch(temp,target);
        if(targetRow >= 0)
            return true;
        else if(targetRow == -1)
            return false;
        else
            targetRow = -targetRow - 2;
        
        if(binarySearch(matrix[targetRow],target) > 0)
            return true;
        else
            return false;
            
    }
    public int binarySearch(int[] arr,int target){
        int low = 0,high = arr.length - 1;
        while(low <= high){
            int center = (low + high) / 2;
            if(arr[center] == target)
                return center;
            else if(arr[center] < target){
                low = center + 1;
            }
            else{
                high = center - 1;
            }
        }
        //如果不存在，返回target應在陣列中的位置，為了跟return center可能為0區分開，這裡多減去1
        return - low - 1;
        
        
    }
}