Search a 2D Matrix

Search a 2D Matrix

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.

解题技巧：

先用二分搜索法找到目标所在的行，再用二分搜索法得到目标所在的列。如果找到目标，则返回true；否则，返回false。时间复杂度为o(logm+logn)

代码：

bool searchMatrix(vector< vector<int> >& matrix, int target){    int bg1, ed1, mid1, bg2, ed2, mid2;         if(matrix.size() == 0) return false;        bg1 = 0, ed1 = matrix.size();    //行    bg2 = 0, ed2 = matrix[0].size(); //列       while(bg1 <= ed1)    {        mid1 = (bg1 + ed1)/2;        if(matrix[mid1][0] == target) return true;        else if(matrix[mid1][0] > target) ed1 = mid1 - 1;        else if(matrix[mid1][0] < target)        {            if(mid1 + 1 >= matrix.size()) break;            else            {                if(target < matrix[mid1+1][0]) break;                else bg1 = mid1 + 1;            }        }    }    while(bg2 <= ed2)    {         mid2 = (bg2 + ed2)/2;         if(matrix[mid1][mid2] == target) return true;         else if(matrix[mid1][mid2] > target) ed2 = mid2 - 1;         else bg2 = mid2 + 1;    }    return false;}