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;

    }