Search a 2D Matrix

题目：

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.

 

解题思路：利用两次二分查找法。因为所给矩阵第一列也是升序排列的，所以可以先对第一列进行二分查找，锁定该元素所在行数，然后再对列进行二分查找，即可判断target是否存在。这个的算法时间复杂度是O(log(rows)+log(columns))。 

 

代码：

 1 public class Solution {

 2     public boolean searchMatrix(int[][] matrix, int target) 

 3     {

 4         if(matrix.length==0||matrix[0].length==0||matrix==null) return false;

 5         

 6         int low=0;

 7         int high=matrix.length-1;

 8         

 9         while(low<=high)

10         {

11             int mid=(high+low)/2;

12             if(target>matrix[mid][0])

13             {

14                 low=mid+1;

15             }

16             else if (target<matrix[mid][0])

17             {

18                 high=mid-1;

19             }

20             else return true;

21         }

22         

23         int row=high; //当从while中跳出时，low指向的值肯定比target大，而high指向的值肯定比target小 24         

25         if(row<0) return false; 

26         

27         low=0;

28         high=matrix[0].length-1;

29         while(low<=high)

30         {

31             int mid=(high+low)/2;

32             if(target>matrix[row][mid])

33             {

34                 low=mid+1;

35             }

36             else if (target<matrix[row][mid])

37             {

38                 high=mid-1;

39             }

40             else return true;

41         }

42         

43         return false;

44     }

45 }

 

reference： http://blog.csdn.net/linhuanmars/article/details/24216235

http://www.cnblogs.com/springfor/p/3857959.html