[LeetCode] Search a 2D Matrix

Problem:

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.

Solution1:O(m+n)

public class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0)
            return false;
            
        int x = 0;
        int y = matrix[0].length -1;
        
        while(x >=0 && y >=0 && x< matrix.length){
            if(matrix[x][y] == target)
                return true;
            else if(matrix[x][y] > target){
                y--;
            }else{
                x++;
            }    
        }
        return false;   
    }
}

Solution2: O(log(m)+log(n))

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

Key Concept: From right top to search