Leetcode-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.

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Analysis:

This is a double binary search problem. First, we use binary search to determine the row in which the target is; Second, we use binary search to search the determined row.

Solution:

 1 public class Solution {
 2     public boolean searchMatrix(int[][] matrix, int target) {
 3         int xLen = matrix.length;
 4         if (xLen==0) return false;
 5         int yLen = matrix[0].length;
 6         if (yLen==0) return false;
 7 
 8         //Determine row.
 9         int x = 0;
10         int y = xLen-1;
11         int k = -1;
12         int targetRow = -1;
13         while (targetRow==-1){
14             if (y<x) return false;
15 
16             k = x + (y-x)/2;
17             if (target<matrix[k][0]){
18                 y = k-1;
19                 continue; 
20             } else if (target>matrix[k][yLen-1]){
21                 x = k+1;
22                 continue;
23             } else {
24                 targetRow = k;
25                 continue;
26             }
27         }
28 
29         //Search the target element in the targetRow
30         x = 0;
31         y = yLen-1;
32         k = -1;
33         boolean findTarget = false;
34         while (!findTarget){
35             if (y<x) {
36                 findTarget = false;
37                 break;
38             }        
39          
40             k=x+(y-x)/2;
41             if (target==matrix[targetRow][k]){
42                 findTarget = true;
43                 break;
44             } else if (target<matrix[targetRow][k]){
45                 y = k-1;
46                 continue;
47             } else {
48                 x = k+1;
49                 continue;
50             }
51         }
52         return findTarget;
53     }
54 }