Question：

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.

Algorithm：

先对第一列二分查找，然后再对该行二分查找

Accepted Code：

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        int M=matrix.size();
        int N=matrix[0].size();
        int top=0;
        int bottom=M-1;
        int mid1=0;
        if(matrix[0][0]==target)
            return true;
        while(top<=bottom)
        {
            mid1=top+(bottom-top)/2;
            if(matrix[mid1][0]==target)
                return true;
            else if(matrix[mid1][0]<target)
                top=mid1+1;
            else
                bottom=mid1-1;
        }
        if(matrix[mid1][0]>target)
        {
            if(mid1>0)
                mid1--;
            else
                return false;
        }
        int left=0;
        int right=N-1;
        while(left<=right)
        {
            int mid2=left+(right-left)/2;
            if(matrix[mid1][mid2]==target)
                return true;
            else if(matrix[mid1][mid2]<target)
                left=mid2+1;
            else
                right=mid2-1;
        }
        return false;
    }
};