> For the complete documentation index, see [llms.txt](https://liuyang89116.gitbook.io/my-leetcode-book/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://liuyang89116.gitbook.io/my-leetcode-book/chapter_3_binary_search/problem_240_search_a_2d_matrix_ii.md).

# Problem 240: Search a 2D Matrix II

> <https://leetcode.com/problems/search-a-2d-matrix-ii/>

## 思路

![](https://1241747088-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-Lpv9LvBSlFaukf_ALqh%2F-Lpv9NPw3Ji1X5Vk8CES%2F-Lpv9vKGNvESa1lydnL8%2F2DMatrixII_1.jpg?generation=1569729532197148\&alt=media)

首先要想清楚出发点。**为什么我们从左下角一路出发？**

因为这个点十分关键，他起到了四分法的作用。相当于他在角落里，进可攻退可守。比他大，只能往右走，比他小，只能往左走。

```java
public class Solution {
    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 n = matrix.length;
        int m = matrix[0].length;
        int x = n - 1;
        int y = 0;
        while (x >= 0 && y < m) {
            if (target == matrix[x][y]) {
                return true;
            } else if (target > matrix[x][y]) {
                y++;
            } else {
                x--;
            }

        }

        return false;
    }
}
```

## 易错点

1. 找好定位点

   ```java
   int n = matrix.length;
   int m = matrix[0].length;
   int x = n - 1;
   int y = 0;
   ```