LeetCode Longest Increasing Path in a Matrix

原题链接在这里：https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

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.

dp[i][j] 表示 当前i, j 位置能都到的最大距离。

dp[i][j] 是通过 dfs 来选的，初始dp[i][j] 都是0, 终止条件是若是走到了一个不是0的位置，那么直接返回dp[x][y]. 可以避免重复计算. 若是dp[x][y]已经有值，说明这个点4个方向的最大值已经找到, 找到dp[x][y]的路径必定是比matrix[x][y]小的，直接返回dp[x][y] 再加上 1 就是之前位置的最大延展长度了。

若是当前位置是0, 就从上下左右四个方向dfs, 若是过了边界或者新位置matrix[x][j] <= 老位置matrix[i][j], 直接跳过continue.

不然len = 1 + dfs. 取四个方向最大的len作为dp[i][j].

Time Complexity: 对于每一个点都做dfs, dfs O(m*n). 所以一共 O(m*n * m*n) = O(m^2 * n^2).

Space: O(m*n).用了dp array.

AC Java:

 1 public class Solution {
 2     final int [][] fourD = {{-1, 0}, {1,0}, {0,-1}, {0,1}};
 3     
 4     public int longestIncreasingPath(int[][] matrix) {
 5         if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
 6             return 0;
 7         }
 8         int max = 1;
 9         int m = matrix.length;
10         int n = matrix[0].length;
11         int [][] dp = new int[m][n];
12         for(int i = 0; i<m; i++){
13             for(int j = 0; j<n; j++){
14                 dp[i][j] = dfs(matrix, i, j, m, n, dp);
15                 max = Math.max(max, dp[i][j]);
16             }
17         }
18         return max;
19     }
20     
21     private int dfs(int [][] matrix, int i, int j, int m, int n, int [][] dp){
22         if(dp[i][j] != 0){
23             return dp[i][j];
24         }
25         int max = 1;
26         for(int k = 0; k<fourD.length; k++){
27             int x = i+fourD[k][0];
28             int y = j+fourD[k][1];
29             if(x<0 || x>=m || y<0 || y>=n || matrix[x][y] <= matrix[i][j]){
30                 continue;
31             }
32             int len = 1 + dfs(matrix, x, y, m, n, dp);
33             max = Math.max(max, len);
34         }
35         dp[i][j] = max;
36         return dp[i][j];
37     }
38 }