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.

已排序的矩阵，先对比每行最后一个值，若比目标值大，则目标值肯定在该行搜索，而后使用二分搜索。

class Solution {
public:
    bool searchMatrix(vector<vector<int>>& matrix, int target) {
        if(matrix.size() == 0 || matrix[0].size() == 0)
            return false;
        int i = 0,j = matrix[0].size() - 1;
        while(i < matrix.size()){
            if(matrix[i][j] == target)
                return true;
            if(matrix[i][j] < target)
                i++;
            else{
                int k = 0;
                j --;
                while(k <= j){
                    int mid = k + ((j - k) >> 1);
                    if(matrix[i][j] == target || matrix[i][k] == target || matrix[i][mid] == target)
                        return true;
                    if(matrix[i][mid] > target)
                        j = mid - 1;
                    else
                        k = mid + 1;
                }
                return false;
            }
        }
        return false;
    }
};