Matrix Search

Write an efficient algorithm that searches for a value in an m x n matrix.

This matrix has the following properties:

  1. Integers in each row are sorted from left to right.
  2. The first integer of each row is greater than or equal to the last integer of the previous row.
Example:

Consider the following matrix:

[
  [1,   3,  5,  7],
  [10, 11, 16, 20],
  [23, 30, 34, 50]
]
Given target = 3, return 1 ( 1 corresponds to true )

Return 0 / 1 ( 0 if the element is not present, 1 if the element is present ) for this problem

思路：

把matrix看做array就行，用mid / n 得到x index，mid % n得到y index。

Solution:

Time: O(logn)
Space: O(1)

public class Solution {
    public int searchMatrix(ArrayList<ArrayList<Integer>> a, int b) {
        //     0 1 2  3
        //     - - -  -
        //  0 |0 1 2  3
        //  1 |4 5 6  7
        //  2 |8 9 10 11
        int m = a.size();
        int n = a.get(0).size();
        int size = m * n;
        int left = 0;
        int right = size - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            int x = mid / n;
            int y = mid % n;
            if (a.get(x).get(y) == b) {
                return 1;
            } else if (a.get(x).get(y) < b) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return 0;
    }
}