【leetcode】Search a 2D Matrix

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.

 

最简单的想法，转化成一维的，再用二分法：

 

 1 class Solution {

 2 public:

 3     bool searchMatrix(vector<vector<int> > &matrix, int target) {

 4        

 5         int row=matrix.size();

 6         int col=matrix[0].size();

 7         vector<int> m(row*col);

 8        

 9         for(int i=0;i<row;i++)

10         {

11             for(int j=0;j<col;j++)

12             {

13                 m[i*col+j]=matrix[i][j];

14             }

15         }

16        

17         int left=0;

18         int right=m.size()-1;

19         int mid;

20         while(left<=right)

21         {

22             mid=(left+right)/2;

23             if(m[mid]>target)

24                 right=mid-1;

25             else if(m[mid]<target)

26                 left=mid+1;

27             else

28                 return true;

29         }

30         return false;

31        

32     }

33 };

 

 

 

第二种思路，直接使用二分法

 

把一维数字转化为二维的坐标的方法：

第n个元素，在n/col行，n%col列

 

 1 class Solution {

 2 public:

 3     bool searchMatrix(vector<vector<int> > &matrix, int target) {

 4        

 5         int row=matrix.size();

 6         int col=matrix[0].size();

 7        

 8         int left=0;

 9         int right=row*col-1;

10         int mid;

11         int m;

12         while(left<=right)

13         {

14             mid=(left+right)/2;

15             m=matrix[mid/col][mid%col];

16             if(m>target)

17                 right=mid-1;

18             else if(m<target)

19                 left=mid+1;

20             else

21                 return true;

22         }

23         return false;          

24     }

25 };