74. 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.

Solution:

This is a binary search problem.

We map the two dimensional coordinates to one dimensional number.

Thus, the time complexity can be optimized to O(log(mn)).



Code:

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) {
            return false;
        }
        if (matrix[0] == null || matrix[0].length == 0) {
            return false;
        }
        int m = matrix.length;
        int n = matrix[0].length;
        int start = 0;
        int end = m * n - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (matrix[mid / n][mid % n] <= target) {
                start = mid;
            }
            else {
                end = mid;
            }
        }
        if (matrix[end / n][end % n] == target) {
            return true;
        }
        if (matrix[start / n][start % n] == target) {
            return true;
        }
        return false;
    }
}