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.
int dx[] = { 1 , -1, 0 , 0 }; int dy[] = { 0 , 0 , 1 , -1 }; class Solution { public: int dfs(int x, int y, const int &m,const int &n,vector<vector<int>>& matrix, vector<vector<int>>& dis) { if (dis[x][y]) return dis[x][y]; for (int i = 0; i < 4; i++) { int nx = x + dx[i]; int ny = y + dy[i]; if (nx >= 0 && ny >= 0 && nx < m && ny < n && matrix[nx][ny] > matrix[x][y]) { dis[x][y] = max(dis[x][y], dfs(nx, ny, m, n, matrix, dis)); } } return ++dis[x][y]; } int longestIncreasingPath(vector<vector<int>>& matrix) { if (!matrix.size()) return 0; int m = matrix.size(), n = matrix[0].size(); vector<vector<int> > dis(m, vector<int>(n, 0)); int ans = 0; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { ans = max(ans, dfs( i, j, m, n, matrix, dis)); } } return ans; } };
