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[Leetcode] search a 2d matrix 搜索二维矩阵

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, returntrue.

 题意:在一个二维矩阵中,查询一个数是否存在。数组:1)每行从左到右从下到大排好;2)行首元素大于上一行的最后一个元素;

思路:常规思路:先遍历行找到元素所可能在的行,然后遍历列,判断是否在在该行中,时间复杂度O(n+m);二分查找版本一:是对常规思路的升级,先查找行 ,再查找列,但这时使用的查找的方法不是从头到尾的遍历,是二分查找,值得注意的是查找完行以后的返回值,时间复杂度O{logn+logm)二分查找版本二:因为矩阵数排列的特性,可以看成一个排列好的一维数组[0, n*m],可以针对整个二维矩阵进行二分查找,时间复杂还是O(log(n*m)),这里的难点是,如何将二维数组的下标和一维数组之间进行转换。

方法一:

 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         int subRow = 0;
 8         if (row == 0 || col == 0)    return false;
 9 
10         //寻找行
11         if (matrix[row - 1][0] <= target)   //最后一行,特殊处理
12             subRow = row - 1;
13         else
14         {
15             for (int i = 0; i<row - 1; ++i)
16             {
17 
18                 if ((matrix[i][0] <= target) && (matrix[i + 1][0]>target))
19                 {
20                     subRow = i;
21                     break;
22                 }
23             }
24         }
25 
26         //查找列
27         for (int j = 0; j<col; ++j)
28         {
29             if (matrix[subRow][j] == target)
30                 return true;
31         }
32         return false;
33     }
34 };

 

方法二:如下:

 1 // Two binary search
 2 class Solution {
 3 public:
 4     bool searchMatrix(vector<vector<int> > &matrix, int target) 
 5     {
 6         int row=matrix.size();
 7         int col=matrix[0].size();
 8         if (row==0 || col==0) return false;
 9         if (target < matrix[0][0] || target > matrix[row-1][col-1]) return false;
10 
11         //查找行
12         int lo = 0, hi = row - 1;
13         while (lo <= hi) 
14         {
15             int mid = (lo+hi) / 2;
16             if (matrix[mid][0] == target) 
17                 return true;
18             else if (matrix[mid][0] < target) 
19                 lo = mid + 1;
20             else 
21                 hi = mid - 1;
22         }
23         int tmp = hi;       //特别注意
24         //查找该行   
25         lo = 0;
26         hi = col - 1;
27         while (lo <= hi) 
28         {
29             int mid = (lo+hi) / 2;
30             if (matrix[tmp][mid] == target) 
31                 return true;
32             else if (matrix[tmp][mid] < target) 
33                 lo = mid + 1;
34             else 
35                 hi = mid - 1;
36         }
37         return false;
38     }
39 };

 

方法三:

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target)
    {
        int row = matrix.size();
        int col = matrix[0].size();

        if(row==0||col==0)  return false;
        if(matrix[0][0]>target||target>matrix[row-1][col-1])    return false;    //加与不加都行

        int lo=0,hi=row*col-1;
        while(lo<=hi)
        {
            int mid=(lo+hi)/2;
            int i=mid/col;
            int j=mid%col;
            if(target==matrix[i][j])
                return true;
            else if(target>matrix[i][j])
                lo=mid+1;
            else
                hi=mid-1;
        }
        return false;
    }
};

 

[Leetcode] search a 2d matrix 搜索二维矩阵