Search a 2D Matrix

Total Accepted: 43629 Total Submissions: 138231

 

 

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.

常规思路, 多次二分

bool searchMatrix(vector
     
      
       >& matrix, int target) {        if(matrix.empty())            return false;        vector
       
         firstCol;        for(auto v : matrix)        firstCol.push_back(v[0]);        if(binary_search(firstCol.begin(), firstCol.end(), target))            return true;         vector
        
         ::iterator iter = lower_bound(firstCol.begin(), firstCol.end(), target);        int low = iter - firstCol.begin();        for(int i = 0; i < low; i++)        {            if(binary_search(matrix[i].begin(), matrix[i].end(), target))                return true;        }        return false;    }
        
       
      
     

 

投机取巧,存在潜在的bug

 

bool searchMatrix(vector
     
      
        >& matrix, int target) {        int rows = matrix.size();        if (rows == 0)    return false;        int cols = matrix[0].size();        int first = 0, last = rows * cols - 1;        if (target < matrix[0][0] || target > matrix[rows - 1][cols - 1])            return false;        int mid, val;        // binary search        while (first <= last) {            mid = first + (last - first) / 2;            val = matrix[mid / cols][mid % cols];            if (val > target)                last = mid - 1;            else if (val == target)                return true;            else                first = mid + 1;        }        return false;    }