題目描述
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:
Input: nums =
[
[9,9,4],
[6,6,8],
[2,1,1]
]
Output: 4
Explanation: The longest increasing path is [1, 2, 6, 9].
Example 2:
Input: nums =
[
[3,4,5],
[3,2,6],
[2,2,1]
]
Output: 4
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.
代碼
class Solution {
public:
int longestIncreasingPath(vector<vector<int>>& matrix) {
if (matrix.empty() || matrix[0].empty()) return 0;
int n = matrix.size();
int m = matrix[0].size();
vector<vector<int>> res(n, vector<int>(m, 0));
int ans = 0;
for (int i=0; i<n; ++i) {
for (int j=0; j<m; ++j) {
if (res[i][j] == 0) {
ans = max(ans, bfs(i, j, res, matrix));
}
}
}
return ans;
}
vector<int> dirs {1, 0, -1, 0, 1};
int bfs(int x, int y, vector<vector<int>>& res, vector<vector<int>>& matrix) {
if (res[x][y] != 0) return res[x][y];
int ans = 0;
for (int i=0; i<4; ++i) {
int nx = x + dirs[i];
int ny = y + dirs[i+1];
if (nx < 0 || nx >= res.size() || ny < 0 || ny >= res[0].size()) {
continue;
}
if (matrix[nx][ny] >= matrix[x][y]) continue;
ans = max(ans, bfs(nx, ny, res, matrix));
}
res[x][y] = ans+1;
return res[x][y];
}
};