Google+ Followers

Sunday, March 17, 2013

[LeetCode] Search a 2D Matrix, Solution


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.
» Solve this problem

[Thoughts]
做两次二分就好了,首先二分第一列,找出target所在的行,然后二分该行。

[Code]
1:       bool searchMatrix(vector<vector<int> > &matrix, int target) {  
2:            int row = matrix.size();  
3:            if(row ==0) return false;  
4:            int col = matrix[0].size();  
5:            if(col ==0) return false;      
6:            if(target< matrix[0][0]) return false;  
7:            int start = 0, end = row-1;  
8:            while(start<= end)  
9:            {  
10:                 int mid = (start+end)/2;  
11:                 if(matrix[mid][0] == target)  
12:                      return true;  
13:                 else if(matrix[mid][0] < target)  
14:                      start = mid+1;  
15:                 else  
16:                      end = mid-1;  
17:            }  
18:            int targetRow = end;  
19:            start =0;  
20:            end = col-1;  
21:            while(start <=end)  
22:            {  
23:                 int mid = (start+end)/2;  
24:                 if(matrix[targetRow][mid] == target)  
25:                      return true;  
26:                 else if(matrix[targetRow][mid] < target)  
27:                      start = mid+1;  
28:                 else  
29:                      end = mid-1;  
30:            }  
31:            return false;  
32:       }  

注意,
二分的条件应该是(start<=end), 而不是(start<end)。

Post a Comment