Leetcode-Search a 2D Matrix

LiBlog發表於2014-11-16
LeetCode

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.

Have you met this question in a real interview?
 
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 }

 

相關文章