Leetcode 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 method:

1 

The same question for the binary search., for this question, the middle element is int the middle of the array. We can think the two-dimensional array is a sorted array, Time complexity is O(log(m * n))

2

From the right up element and move until to the left end element. This may be called chain search for the 2-dimension sorted array. O(m  *   n)

public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if(matrix==null || matrix.length==0 || 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<=end){
int mid=(start+end)/2;
int midX=mid/n;
int midY=mid%n;

if(matrix[midX][midY]==target)
return true;

if(matrix[midX][midY]<target){
start=mid+1;
}else{
end=mid-1;
}
}

return false;
}
}