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 4The longest increasing path is [1, 2, 6, 9].Example 2:nums = [  [3,4,5],  [3,2,6],  [2,2,1]]Return 4The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

DFS + DP:

use a two dimensional matrix dp[i][j] to store the length of the longest increasing path starting at matrix[i][j]

transferring function is: dp[i][j] = max(dp[i][j], dp[x][y] + 1), where dp[x][y] is its neighbor with matrix[x][y] > matrix[i][j]

Note:

1. Use matrix[x][y] > matrix[i][j] so we don't need a visited[m][n] array
2. The key is to cache the distance because it's highly possible to revisit a cell

 

Follow Up: How to get the actual longest increasing path

我的想法：类似, 除了一个dp[i][j]记录longest length以外，另外再用一个matrix pre[i][j]记录(i,j)longest increasing path上一跳位置, 并用一个variable记录最后最长的path的起始位置（第19行每次res更新时更新）。最后通过这个起始位置沿着一个一个上一跳位置，可以求出path

1 public class Solution { 2     int[][] dp; 3     int[][] directions = new int[][]{
   {-1,0},{1,0},{0,-1},{0,1}}; 4     int m; 5     int n; 6  7     public int longestIncreasingPath(int[][] matrix) { 8         if (matrix==null || matrix.length==0 || matrix[0].length==0) return 0; 9         m = matrix.length;10         n = matrix[0].length;11         dp = new int[m][n];12 13         int result = 0;14 15         for (int i=0; i
     
      =m || y>=n || matrix[x][y]<=matrix[i][j]) continue;32             dp[i][j] = Math.max(dp[i][j], DFS(x, y, matrix)+1);33         }34         return dp[i][j];35     }36 }