Leetcode: Search a 2D Matrix

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<vector<int>>& matrix, int target) {
        if(matrix.empty())
            return false;
        vector<int> firstCol;
        for(auto v : matrix)
        firstCol.push_back(v[0]);
        if(binary_search(firstCol.begin(), firstCol.end(), target))
            return true; 
        vector<int>::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<vector<int> >& 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;
    }