> For the complete documentation index, see [llms.txt](https://junnie.gitbook.io/nine-chapter/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://junnie.gitbook.io/nine-chapter/1.binary-search/28search-a-2d-matrix.md).

# 28.Search a 2D Matrix

## 1.Description(Easy)

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.

**Example**

Consider the following matrix:

```
[
    [1, 3, 5, 7],
    [10, 11, 16, 20],
    [23, 30, 34, 50]
]
```

Given`target = 3`, return`true`.

[**Challenge**](http://lintcode.com/en/problem/search-a-2d-matrix/#challenge)

O(log(n) + log(m)) time

## 2.Code

把他们变成一维的来做，注意矩阵的表示方法。

```
public boolean searchMatrix(int[][] matrix, int target) {
      if(matrix==null || matrix.length==0){
            return false;
        }
        if(matrix[0]==null || matrix[0].length==0){
            return false;
        }
        int row=matrix.length;
        int column=matrix[0].length;
        int start=0;
        int end=row*column-1;
        while(start+1<end){
            int mid=start+(end-start)/2;
            if(matrix[mid/column][mid%column]==target){
                return true;
            }
            else if(matrix[mid/column][mid%column]>target){
                end=mid;
            }
            else{
                start=mid;
            }
        }
        if(matrix[start/column][start%column]==target){
            return true;
        }
        if(matrix[end/column][end%column]==target){
            return true;
        }
        return false;        
    }
```