Leetcode-329: 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.
Credits:

Special thanks to @dietpepsi for adding this problem and creating all test cases.

给了一个矩阵，求这个矩阵中的最长递增路径。（路径不允许走对角线。）

思路：这种求最长路径的题第一想到的都应该是深度优先搜索（DFS），可以用一个同样大小的矩阵标记以各点为起点的最长递增路径，所以也有动态规划的思想在里面。如果matrix[i][j]周围的数都比ta小的话，那么以matrix[i][j]为起点的最长递增路径长度为1；否则找到matrix[i][j]周围长度最长的那个点(ii, jj)，以matrix[ii][jj]为起点的最长递增路径长度加1即为matrix[i][j]的路径长度。

class Solution {
private int[][] longestPath;
private int[][] direct = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};

public int longestIncreasingPath(int[][] matrix) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
return 0;
longestPath = new int[matrix.length][matrix[0].length];

int max = 0;
for (int i = 0; i < matrix.length; ++i) {
for (int j = 0; j < matrix[0].length; ++j) {
max = Math.max(max, dfs(matrix, i, j));
}
}
return max;
}

private int dfs(int[][] matrix, int i, int j) {
if (longestPath[i][j] != 0)
return longestPath[i][j];

int max = 1;
for (
894d
int[] d : direct) {
int ii = i + d[0], jj = j + d[1];
if (!(ii < 0 || jj < 0 || ii >= matrix.length || jj >= matrix[0].length) && matrix[ii][jj] > matrix[i][j])
max = Math.max(max, 1 + dfs(matrix, ii, jj));
}
longestPath[i][j] = max;
return max;
}
}