LeetCode: Search a 2D Matrix 解题报告

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.

SOLUTION 1:

采用经典的二分法模板

 1 while (left <= right) {
 2             int mid = left + (right - left) / 2;
 3             
 4          
 5             int n = matrix[row][col];
 6             
 7             if (n == target) {
 8                 ...
 9             } else if (n < target) {
10                 left = mid + 1;
11             } else {
12                 right = mid - 1;
13             }
14         }

 

 1 public class Solution {
 2     public boolean searchMatrix(int[][] matrix, int target) {
 3         if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
 4             return false;
 5         }
 6         
 7         int rows = matrix.length;
 8         int cols = matrix[0].length;
 9         
10         int num = rows * cols;
11         
12         int left = 0;
13         int right = num - 1;
14         
15         while (left <= right) {
16             int mid = left + (right - left) / 2;
17             
18             int row = mid / cols;
19             int col = mid % cols;
20             
21             int n = matrix[row][col];
22             
23             if (n == target) {
24                 return true;
25             } else if (n < target) {
26                 left = mid + 1;
27             } else {
28                 right = mid - 1;
29             }
30         }
31         
32         return false;        
33     }
34 }
View Code

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/divide2/SearchMatrix.java

posted on 2014-12-31 23:50  Yu's Garden  阅读(994)  评论(2)  编辑  收藏  举报

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