可能重复：   Given二维数组排序从左至右，从上到下递增的顺序，什么是搜索目标数量的最佳方法是什么？

下面被要求在谷歌的采访：

The following was asked in a Google interview:

您将得到一个二维数组存储整数，垂直和水平方向排列。

You are given a 2D array storing integers, sorted vertically and horizontally.

写一个方法，作为输入一个整数，输出布尔话说整数是否在数组中

Write a method that takes as input an integer and outputs a bool saying whether or not the integer is in the array.

什么是做到这一点的最好方法是什么？什么是它的时间复杂度？

What is the best way to do this? And what is its time complexity?

推荐答案

我要问的细节意味着什么是启动来分类的垂直和水平

I would start by asking details about what it means to be "sorted vertically and horizontally"

如果该矩阵排序的方式，每一行的最后一个元素是小于下一行的第一个元素，可以运行在第一列上的二进制搜索，找出在什么行这个数字，并然后运行该行上的另一个二进制搜索。这个算法将需要为O（log C +登录R）的时间，其中C和R，行和列的分别的数目。使用对数的特性，可以写为○（日志（C * R））的，这是相同的为O（log N）时，如果N是数组中元素的数量。这是几乎相同处理阵一维和在其上运行的二进制搜索。

If the matrix is sorted in a way that the last element of each row is less than the first element of the next row, you can run a binary search on the first column to find out in what row that number is, and then run another binary search on the row. This algorithm will take O(log C + log R) time, where C and R are, respectively the number of rows and columns. Using a property of the logarithm, one can write that as O(log(C*R)), which is the same as O(log N), if N is the number of elements in the array. This is almost the same as treating the array as 1D and running a binary search on it.

但是，基质可能的方式，每一行的最后一个元素不大于下一行的第一个元素少进行排序：

But the matrix could be sorted in a way that the last element of each row is not less than the first element of the next row:

1 2 3 4 5 6 7 8  9
2 3 4 5 6 7 8 9  10
3 4 5 6 7 8 9 10 11

在这种情况下，你可以同时运行某种水平的垂直二进制搜索的：

In this case, you could run some sort of horizontal an vertical binary search simultaneously:

此方法也对数上的元素的数量。

This method is also logarithmic on the number of elements.