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.
详见：https://leetcode.com/problems/longest-increasing-path-in-a-matrix/description/
C++：
class Solution {
public:
vector<vector<int>> dirs = {{0, -1}, {-1, 0}, {0, 1}, {1, 0}};
int longestIncreasingPath(vector<vector<int>>& matrix)
{
if (matrix.empty() || matrix[0].empty())
{
return 0;
}
int res = 1, m = matrix.size(), n = matrix[0].size();
vector<vector<int>> dp(m, vector<int>(n, 0));
for (int i = 0; i < m; ++i)
{
for (int j = 0; j < n; ++j)
{
res = max(res, dfs(matrix, dp, i, j));
}
}
return res;
}
int dfs(vector<vector<int>> &matrix, vector<vector<int>> &dp, int i, int j)
{
if (dp[i][j])
{
return dp[i][j];
}
int mx = 1, m = matrix.size(), n = matrix[0].size();
for (auto a : dirs)
{
int x = i + a[0], y = j + a[1];
if (x < 0 || x >= m || y < 0 || y >= n || matrix[x][y] <= matrix[i][j])
{
continue;
}
int len = 1 + dfs(matrix, dp, x, y);
mx = max(mx, len);
}
dp[i][j] = mx;
return mx;
}
};
参考：https://www.cnblogs.com/grandyang/p/5148030.html
329 Longest Increasing Path in a Matrix 矩阵中的最长递增路径
原文：https://www.cnblogs.com/xidian2014/p/8833035.html