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.

矩陣中的最長遞增路徑。題意是給一個二維矩陣，找出最長的遞增路徑。例子應該很好地解釋了什麼叫做最長的遞增路徑。

這道題也是偏flood fill那一類的題，同時因為是找最長路徑，所以用DFS做更好。因為路徑是連續的且只是往上下左右四個方向，所以需要再建立一個同樣size的二維陣列來記錄遍歷的結果，這樣就不需要遍歷整個input陣列了，因為input陣列中比如數字為9的座標，實際是沒什麼掃描的意義的，因為數字非常大了，而他的鄰居搞不好都比他本身小。具體做法還是應用DFS的模板，但是在設計dfs函式的時候記得加一個額外的二維陣列記錄結果，也加一個prev值，用來和當前座標上的值作比較。其餘的部分可參見程式碼。

時間O(mn)

空間O(mn) - 額外陣列快取結果

Java實現

 1 class Solution {
 
 2     public int longestIncreasingPath(int[][] matrix) {
 3         if (matrix == null || matrix.length == 0) {
 4             return 0;
 5         }
 6         int rows = matrix.length;
 7         int cols = matrix[0].length;
 8         int[][] dp = new int[rows][cols];
 9         int res = 0;
10         for (int i = 0; i < rows; i++) {
11             for (int j = 0; j < cols; j++) {
12                 if (dp[i][j] == 0) {
13                     dfs(matrix, i, j, dp, Integer.MIN_VALUE);
14                     res = Math.max(res, dp[i][j]);
15                 }
16             }
17         }
18         return res;
19     }
20 
21     private int dfs(int[][] matrix, int row, int col, int[][] dp, int prev) {
22         if (row > matrix.length - 1 || row < 0 || col > matrix[0].length - 1 || col < 0 || matrix[row][col] <= prev) {
23             return 0;
24         }
25         if (dp[row][col] != 0) {
26             return dp[row][col];
27         }
28         int left = dfs(matrix, row, col - 1, dp, matrix[row][col]);
29         int right = dfs(matrix, row, col + 1, dp, matrix[row][col]);
30         int up = dfs(matrix, row - 1, col, dp, matrix[row][col]);
31         int down = dfs(matrix, row + 1, col, dp, matrix[row][col]);
32         dp[row][col] = Math.max(left, Math.max(right, Math.max(up, down))) + 1;
33         return dp[row][col];
34     }
35 }

flood fill題型總結

LeetCode 題目總結