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.

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给定一个矩阵，求矩阵中最长的升序（严格升序）序列的长度。序列可以上下左右延伸。一开始想到用dfs的方法，但问题是如果对每一个点都进行dfs，肯定是会超时的。然后又瞄了一眼tags，看到有memorization，想到了可以保存结果，避免重复dfs。想法是对一个点dfs，如果这个点是有记录的，直接返回记录值，不然的话上下左右四个方向dfs，然后得到四个值，取最大的加1代表从这个点开始的最长升序路径的长，记录好。这样dfs过程中经过的每个点都会记录好，大大减少了计算次数。

代码：

class Solution
{
public:
	int longestIncreasingPath(vector<vector<int>>& matrix) 
	{
		m = matrix.size(); if(m == 0) return 0;
		n = matrix[0].size(); if(n == 0) return 0;
		int res = 1;

		records.resize(m, vector<int>(n, 0));

		for(int i = 0; i < m; ++i)
		{
			for(int j = 0; j < n; ++j)
			{
				res = max(res, dfs(matrix, i, j));
			}
		}
		return res;
	}
private:
	int m, n;
	vector<vector<int> >records;
	int dfs(vector<vector<int> > &matrix, int i, int j)
	{
		if(records[i][j]) return records[i][j];
		if(i == -1 || i == m || j == -1 || j == n) return 0;	
		int up = 0, down = 0, left = 0, right = 0;
		if(i > 0 && matrix[i-1][j] > matrix[i][j]) up = dfs(matrix, i-1, j);
		if(i < m-1 && matrix[i+1][j] > matrix[i][j]) down = dfs(matrix, i+1, j);
		if(j > 0 && matrix[i][j-1] > matrix[i][j]) left = dfs(matrix, i, j-1);
		if(j < n-1 && matrix[i][j+1] > matrix[i][j]) right = dfs(matrix, i, j+1);
		records[i][j] = max(max(up, down), max(left, right)) + 1;
		return records[i][j];
	}
};