> For the complete documentation index, see [llms.txt](https://wentao-shao.gitbook.io/leetcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://wentao-shao.gitbook.io/leetcode/graph-search/329.longest-increasing-path-in-a-matrix.md).

# 329.Longest-Increasing-Path-in-a-Matrix

## 329. Longest Increasing Path in a Matrix

## 题目地址

<https://leetcode.com/problems/longest-increasing-path-in-a-matrix/>

## 题目描述

```
Given an integer matrix, find the length of the longest increasing path.

From each cell, you can either move to four directions: left, right, up or down. You may NOT move diagonally or move outside of the boundary (i.e. wrap-around is not allowed).

Example 1:
Input: nums = 
[
  [9,9,4],
  [6,6,8],
  [2,1,1]
] 
Output: 4 
Explanation: The longest increasing path is [1, 2, 6, 9].

Example 2:
Input: nums = 
[
  [3,4,5],
  [3,2,6],
  [2,2,1]
] 
Output: 4 
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
```

## 代码

### Approach #1 DFS + Memoization

```java
class Solution {
  private static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
  private int m, n;
  public int longestIncreasingPath(int[][] matrix) {
        if (matrix.length == 0)        return 0;
    m = matrix.length;
    n = matrix[0].length;
    int[][] cache = new int[m][n];
    int ans = 0;
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        ans = Math.max(ans, dfs(matrix, i, j, cache));
      }
    }
    return ans;
  }

  private int dfs(int[][] matrix, int i, int j, int[][] cache) {
    if (cache[i][j] != 0)        return cache[i][j];
    for (int[] d: dirs) {
      int x = i + d[0];
      int y = j + d[1];
      if (x >= 0 && x < m
         && y >= 0 y < n
         && matrix[x][y] > matrix[i][j]) {

        cache[i][j] = Math.max(cache[i][j], dfs(matrix, x, y, cache));
      }
    }

    return ++cache[i][j];
  }
}
```

### Approach #3 Peeling Onion Confusion

```java
// Topological Sort Based Solution
// An Alternative Solution
public class Solution {
    private static final int[][] dir = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    private int m, n;
    public int longestIncreasingPath(int[][] grid) {
        int m = grid.length;
        if (m == 0) return 0;
        int n = grid[0].length;
        // padding the matrix with zero as boundaries
        // assuming all positive integer, otherwise use INT_MIN as boundaries
        int[][] matrix = new int[m + 2][n + 2];
        for (int i = 0; i < m; ++i)
            System.arraycopy(grid[i], 0, matrix[i + 1], 1, n);

        // calculate outdegrees
        int[][] outdegree = new int[m + 2][n + 2];
        for (int i = 1; i <= m; ++i)
            for (int j = 1; j <= n; ++j)
                for (int[] d: dir)
                    if (matrix[i][j] < matrix[i + d[0]][j + d[1]])
                        outdegree[i][j]++;

        // find leaves who have zero out degree as the initial level
        n += 2;
        m += 2;
        List<int[]> leaves = new ArrayList<>();
        for (int i = 1; i < m - 1; ++i)
            for (int j = 1; j < n - 1; ++j)
                if (outdegree[i][j] == 0) leaves.add(new int[]{i, j});

        // remove leaves level by level in topological order
        int height = 0;
        while (!leaves.isEmpty()) {
            height++;
            List<int[]> newLeaves = new ArrayList<>();
            for (int[] node: leaves) {
                for (int[] d: dir) {
                    int x = node[0] + d[0], y = node[1] + d[1];
                    if (matrix[node[0]][node[1]] > matrix[x][y])
                        if (--outdegree[x][y] == 0)
                            newLeaves.add(new int[]{x, y});
                }
            }
            leaves = newLeaves;
        }
        return height;
    }
}
```

### Appraoch #3 TLE

```java
// Naive DFS Solution
// Time Limit Exceeded
public class Solution {
  private static final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
  private int m, n;

  public int longestIncreasingPath(int[][] matrix) {
      if (matrix.length == 0) return 0;
      m = matrix.length;
      n = matrix[0].length;
      int ans = 0;
      for (int i = 0; i < m; ++i)
          for (int j = 0; j < n; ++j)
              ans = Math.max(ans, dfs(matrix, i, j));
      return ans;
  }

  private int dfs(int[][] matrix, int i, int j) {
      int ans = 0;
      for (int[] d : dirs) {
          int x = i + d[0], y = j + d[1];
          if (0 <= x && x < m && 0 <= y && y < n && matrix[x][y] > matrix[i][j])
              ans = Math.max(ans, dfs(matrix, x, y));
      }
      return ++ans;
  }
}
```


---

# Agent Instructions
This documentation is published with GitBook. GitBook is the documentation platform designed so that both humans and AI agents can read, navigate, and reason over technical content effectively. Learn more at gitbook.com.

## 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, and the optional `goal` query parameter:

```
GET https://wentao-shao.gitbook.io/leetcode/graph-search/329.longest-increasing-path-in-a-matrix.md?ask=<question>&goal=<endgoal>
```

`ask` is the immediate question: it should be specific, self-contained, and written in natural language.
`goal` is optional and describes the broader end goal you are ultimately trying to accomplish on behalf of the user. GitBook uses it to tailor the answer towards what is most useful for that goal.

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.