标签：tin problem path asi long rip direction vector sid
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
标签：tin problem path asi long rip direction vector sid
原文地址：https://www.cnblogs.com/xidian2014/p/8833035.html
