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# 74 Search a 2D Matrix

## 74. [Search a 2D Matrix](https://leetcode.com/problems/search-a-2d-matrix/description/)

## 1. Question

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`.

## 2. Implementation

**(1) Binary Search**

思路: 因为每一行的第一个数肯定比上一行的最后一个数大，所以matrix\[0]\[0]到matrix\[m - 1]\[n - 1]是严格递增的, 可以直接把matrix降成一维数组，对这个一维数组进行二分查找

```java
class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        if (matrix == null || matrix.length == 0) {
            return false;
        }

        int m = matrix.length, n = matrix[0].length;
        int start = 0, end = m * n - 1, mid = 0;

        while (start <= end) {
            mid = start + (end - start) / 2;
            int row = mid / n;
            int col = mid % n;

            if (matrix[row][col] == target) {
                return true;
            }
            else if (matrix[row][col] < target) {
                start = mid + 1;
            }
            else {
                end = mid - 1;
            }
        }
        return false;
    }
}
```

## 3. Time & Space Complexity

时间复杂度O(logmn), m为matrix的行数, row为matrix的列数，空间复杂度O(1)