# 240.Search-a-2D-Matrix-II

## 240. Search a 2D Matrix II

## 题目地址

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

## 题目描述

```
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 in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.

Example:
Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]
Given target = 5, return true.

Given target = 20, return false.
```

## 代码

### Approach #1 Binary Search

Time complexity :`O(log(n!))`

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

    int shoterDim = Math.min(matrix.length, matrix[0].length);
    for (int start = 0; start < shorterDim; start++) {
      boolean verticalFound = binarySearch(matrix, target, start, true);
      boolean horizontalFound = binarySearch(matrix, target, start, false);

      if (verticalFound || horizontalFound) {
        return true;
      }
    }

    return false;
  }

  private boolean binarySearch(int[][] matrix, int target, int start, boolean vertical) {
    int left = start;
    int right = vertical ? matrix[0].length - 1: matrix.length - 1;

    while (left <= right) {
      int mid = (left + right) / 2;
      if (vertical) {
        if (matrix[left][mid] < target) {
          left = mid + 1;
        } else if (matrix[left][mid] > target) {
          right = mid - 1;
        } else {
          return true;
        }
      } else {
        if (matrix[mid][left] < target) {
          left = mid + 1;
        } else if (matrix[mid][left] > target) {
          right = mid - 1;
        } else {
          return true;
        }
      }
    }

    return false;
  }
}
```

### Approach #2 Divide and Conquer

```java
class Solution {
  private int[][] matrix;
  private int target;

  public boolean searchMatrix(int[][] mat, int target) {
    if (matrix == null || matrix.length == 0)        return false;

    matrix = mat;
    this.target = target;

    return searchRec(0, 0, matrix[0].length - 1, matrix.length - 1);
  }

  private boolean searchRec(int left, int up, int right, int down) {
    if (left > right || up > down) {
      return false;
    } else if (target < matrix[up][left] || target > matrix[down][right]) {
      return false;
    }

    int mid = left + (right - left) / 2;
    // Locate `row` such that matrix[row-1][mid] < target < matrix[row][mid]
    int row = up;
    while (row <= down && matrix[row][mid] <= target) {
      if (matrix[row][mid] == target)     return true;
      row++;
    }

    return searchRec(left, row, mid - 1, down)
      || searchRec(mid + 1, up, right, row - 1);
  }
}
```

### Approach #3

Time complexity : O(*n*+*m*)

```java
class Solution {
    public boolean searchMatrix(int[][] matrix, int target) {
        // start our "pointer" in the bottom-left
        int row = matrix.length - 1;
        int col = 0;

        while (row >= 0 && col < matrix[0].length) {
            if (matrix[row][col] > target) {
                row--;
            } else if (matrix[row][col] < target) {
                col++;
            } else { // found it
                return true;
            }
        }

        return false;
    }
}
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/matrix/240.search-a-2d-matrix-ii.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.