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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.
剑指offer上的老题了，矩阵是排好序的，那末我们可以从其中找到规律。
从右上角(0，n)开始扫描，如果target比它大就往下找，如果小就往左侧找。
具体看代码：
class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
if(matrix.size()==0) return false;
int i = matrix.size()-1;
int j = 0;
while(i>=0 && j< matrix[0].size())
{
if(target==matrix[i][j]) return true;//找到
if(target>matrix[i][j]) j++;//如果target大，就往下找
else i--;//反之则往左找
}
return false;
}
};