# search a 2D matrix

Write an efficient algorithm that searches for a value in anmxnmatrix. 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`.

分析

1. 2次二分，先找column,再找row.第一次用column用0，loop row. 最后判断row <= target。第二次row用前面二分结果，loop column.
2. ```
   start = 0, end = row * column - 1;
   int mid = start + (end - start) / 2;
   int number = matrix[mid / column][mid % column];
   ```

答案

```
class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if(matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0)
            return false;
        int n = matrix.length;
        int mm = matrix[0].length;
        //row
        int s = 0, e = n - 1, row = 0;       
        while(s + 1 < e){
            int m = s + (e - s)/2;
            if(matrix[m][0] == target){
                return true;
            }else if(matrix[m][0] < target){
                s = m;
            }else{
                e = m;
            }
        }

        if(matrix[s][0] <= target){
            row = s;
        }
        if(matrix[e][0] <= target){
            row = e;
        }
        //column
        s = 0; e = mm - 1;
        while(s + 1 < e){
            int m = s + (e - s)/2;
            if(matrix[row][m] == target){
                return true;
            }else if(matrix[row][m] < target){
                s = m;
            }else{
                e = m;
            }
        }

        if(matrix[row][s] == target){
            return true;
        }
        if(matrix[row][e] == target){
            return true;
        }

        return false;
    }
}
```
