leetcode 74:Search a 2D Matrix

题目：

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

.
思路：
本题可以用二分法来解。注意到将这个数组从第一行第一个从0开始编号，下一行的第一个接上一行的最后一个，最后一个的序号为n1*n2-1;

那么每个元素按照我们编号的顺序是严格递增的。

所以我们可以对index从0-n1*n2-1用二分法来做，唯一要解决的是将index还原到数组中对应的行号和列号。当然这非常容易；

时间复杂度：O(lg(n1*n2))

实现如下：

class Solution {
public:
bool searchMatrix(vector<vector<int>>& matrix, int target) {
int size1 = matrix.size();
if (size1 < 1) return false;
int size2 = matrix[0].size(),n=size1*size2;
if (size2 == 0) n = size1;
int left = 0, right = n - 1,index=n/2;
while (left <= index && index <= right)
{
int row = index / size2, column = index%size2;
if (matrix[row][column] == target) return true;
if (matrix[row][column] > target)
{
right = index - 1;
index -= (index - left + 1) / 2;
}
else
{
left = index + 1;
index += (right - index + 1) / 2;
}
}
return false;
}
};