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
The longest increasing path is
4
The longest increasing path is
[1, 2, 6, 9]
.
Example 2:
nums = [ [3,4,5], [3,2,6], [2,2,1] ]
Return
The longest increasing path is
Solution:4
The longest increasing path is
[3, 4, 5, 6]
. Moving diagonally is not allowed.This is a very classic DFS + memorialization problem. If we only use the DFS solution, it will end with many repeated calculations. Therefore, for each element in the matrix[i][j], we use a DP array dp[i][j] to denote the length of the maximum increasing path from this point. So along with the DFS, for a point in the matrix, if we've already found the longest increasing path, we don't have to repeatedly compute it again; we just need to return the length, which is dp[i][j].
One trick here is dp[i][j] stores the length of the longest increasing path. That is because the DFS from a point matrix[i][j] can guarantee the longest path from this point. Since we store this value in the dp[i][j], that can guarantee that dp[i][j] is the longest path from the point matrix[i][j].
Code (Java):
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 | public class Solution { private int [] dx = new int []{ 0 , 0 , - 1 , 1 }; private int [] dy = new int []{ 1 , - 1 , 0 , 0 }; public int longestIncreasingPath( int [][] matrix) { if (matrix == null || matrix.length == 0 ) { return 0 ; } int m = matrix.length; int n = matrix[ 0 ].length; int max = 0 ; int [][] dp = new int [m][n]; for ( int i = 0 ; i < m; i++) { for ( int j = 0 ; j < n; j++) { max = Math.max(max, helper(i, j, matrix, dp)); } } return max; } private int helper( int row, int col, int [][] matrix, int [][] dp) { if (dp[row][col] > 0 ) { return dp[row][col]; } int m = matrix.length; int n = matrix[ 0 ].length; int curMax = 0 ; for ( int i = 0 ; i < 4 ; i++) { int x = dx[i] + row; int y = dy[i] + col; if (x >= 0 && x < m && y >= 0 && y < n && matrix[x][y] > matrix[row][col]) { curMax = Math.max(curMax, helper(x, y, matrix, dp)); } } dp[row][col] = curMax + 1 ; return curMax + 1 ; } } |
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