Search a 2D Matrix

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.

Tips:

把2D坐标转化为1D然后用binary search做即可。

Code：

public class Solution {
    /**
     * @param matrix, a list of lists of integers
     * @param target, an integer
     * @return a boolean, indicate whether matrix contains target
     */
    public boolean searchMatrix(int[][] matrix, int target) {
        // write your code here
        if (matrix == null || matrix.length == 0){
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0){
            return false;
        }
        int row = matrix.length;
        int column = matrix[0].length;
        int start = 0;
        int end = row * column -1;

        while(start + 1 < end){
            int mid = start + (end - start)/2;
            int first = mid / column;
            int second = mid % column;
            if (matrix[first][second] == target){
                return true;
            }
            else if (matrix[first][second] < target){
                start = mid;
            }
            else{
                end = mid;
            }
        }

        if (matrix[0][0] == target){
            return true;
        }
        if (matrix[row-1][column-1] == target){
            return true;
        }
        else{
            return false;
        }
    }
}