【leetcode】 Search a 2D Matrix (easy)

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.

 

实际上就是二分搜索， 不难， 一次AC。

注意一下退出条件是 l <= r 就行。有等于。

class Solution {
public:
    bool searchMatrix(vector<vector<int> > &matrix, int target) {
        if(matrix.empty())
            return false;

        int length = matrix.size() * matrix[0].size();
        int l = 0, r = length - 1;

        while(l <= r) //注意 这里包含等于
        {
            int mid = (l + r) / 2;
            int col = mid % matrix[0].size();
            int row = mid / matrix[0].size();
            if(matrix[row][col] < target)
            {
                l = mid + 1;
            }
            else if(matrix[row][col] > target)
            {
                r = mid - 1;
            }
            else
            {
                return true;
            }

        }
        return false;
    }
};