Longest Increasing Path in a Matrix

Posted 2020-08-16 北叶青藤

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:

nums = [
  [9,9,4],
  [6,6,8],
  [2,1,1]
]

Return 4
The longest increasing path is [1, 2, 6, 9].

Example 2:

nums = [
  [3,4,5],
  [3,2,6],
  [2,2,1]
]

Return 4
The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

第一种方法，递归。很明显，时间超时，通不过。

 1 public class Solution {
 2     public int longestIncreasingPath(int[][] matrix) {
 3         int[] max = new int[1];
 4         for (int i = 0; i < matrix.length; i++) {
 5             for (int j = 0; j < matrix[0].length; j++) {
 6                 boolean[][] visited = new boolean[matrix.length][matrix[0].length];
 7                 longestIncreasingPathHelper(matrix, visited, i, j, matrix[i][j] - 1, 0, max);
 8             }
 9         }
10         return max[0];
11     }
12 
13     public void longestIncreasingPathHelper(int[][] matrix, boolean[][] visited, int i, int j, int prev, int size,
14             int[] max) {
15         if (i < 0 || i >= matrix.length || j < 0 || j >= matrix[0].length)    return;
16 
17         if (visited[i][j] == true || matrix[i][j] <= prev) return;
18 
19         visited[i][j] = true;
20         size++;
21         max[0] = Math.max(max[0], size);
22         
23         longestIncreasingPathHelper(matrix, visited, i + 1, j, matrix[i][j], size, max);
24         longestIncreasingPathHelper(matrix, visited, i - 1, j, matrix[i][j], size, max);
25         longestIncreasingPathHelper(matrix, visited, i, j + 1, matrix[i][j], size, max);
26         longestIncreasingPathHelper(matrix, visited, i, j - 1, matrix[i][j], size, max);
27         visited[i][j] = false;
28     }
29 }

第二种方法类似第一种方法，但是我们不会每次都对同一个位置重复计算。对于一个点来讲，它的最长路径是由它周围的点决定的，你可能会认为，它周围的点也是由当前点决定的，这样就会陷入一个死循环的怪圈。其实并没有，因为我们这里有一个条件是路径上的值是递增的，所以我们一定能够找到一个点，它不比周围的值大，这样的话，整个问题就可以解决了。

 1 public class Solution {
 2     public int longestIncreasingPath(int[][] A) {
 3         int res = 0;
 4         if (A == null || A.length == 0 || A[0].length == 0) {
 5             return res;
 6         }
 7         int[][] store = new int[A.length][A[0].length];
 8         for (int i = 0; i < A.length; i++) {
 9             for (int j = 0; j < A[0].length; j++) {
10                 if (store[i][j] == 0) {
11                     res = Math.max(res, dfs(A, store, i, j));
12                 }
13             }
14         }
15         return res;
16     }
17 
18     private int dfs(int[][] a, int[][] store, int i, int j) {
19         if (store[i][j] != 0) {
20             return store[i][j];
21         }
22         int left = 0, right = 0, up = 0, down = 0;
23         if (j + 1 < a[0].length && a[i][j + 1] > a[i][j]) {
24             right = dfs(a, store, i, j + 1);
25         }
26         if (j > 0 && a[i][j - 1] > a[i][j]) {
27             left = dfs(a, store, i, j - 1);
28         }
29         if (i + 1 < a.length && a[i + 1][j] > a[i][j]) {
30             down = dfs(a, store, i + 1, j);
31         }
32         if (i > 0 && a[i - 1][j] > a[i][j]) {
33             up = dfs(a, store, i - 1, j);
34         }
35         store[i][j] = Math.max(Math.max(up, down), Math.max(left, right)) + 1;
36         return store[i][j];
37     }
38 }