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.

Example:

Consider following example:

[
  [1, 3, 5, 7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]

Given target = 3, return true.

Challenge:

O(log(n) + log(m)) time

Solution:

We should treat this matrix as a large array. For a[m][n], we could image it as b[m*n].

Code:

public boolean searchMatrix(int[][] matrix, int target) {
    if (matrix == null || matrix.length == 0) {
        return false;
    }
    int m = matrix.length;
    //it's a matrix, the number of columns is same for each row
    int n = matrix[0].length; 
    int start = 0;
    int end = m * n - 1;
    while (start + 1 < end) {
        int mid = start + (end - start) / 2;
        int r = mid / n;
        int c = mid % n;
        if (matrix[r][c] < target) {
            start = mid;
        } else if (matrix[r][c] > target) {
            end = mid;
        } else {
            return true;
        }
    }
    if (matrix[start / n][start % n] == target) {
        return true;
    }
    if (matrix[end / n][end % n] == target) {
        return true;
    }
    return false;
}