1、题目描述
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.
2、问题描述：
- 在一个矩阵中找到一个最长的严格递增的路径。要求不能对角线走。
3、问题关键：
- DP，能走到当前点，说明比当前点小。
- 状态变量：f[x][y]，表示走到（x， y）点的最长路径。
- 状态转移：f[x][y] = max(f[x][y], dp(a, b) + 1); (a, b)是(x, y)周围的点。表示能够从(a, b) 走到(x, y).
4、C++代码：
class Solution {
public:
int n, m;
vector<vector<int>> f;
int dx[4] = {-1, 0, 1, 0}, dy[4] = {0, 1, 0, -1};
int longestIncreasingPath(vector<vector<int>>& matrix) {
if (matrix.empty()) return 0;
n = matrix.size(), m = matrix[0].size();
f = vector<vector<int>> (n, vector<int>(m, -1));
int res = 0;
for (int i = 0; i < n; i ++)
for (int j = 0; j < m; j ++)
res = max(res, dp(i, j, matrix));
return res;
}
int dp(int x, int y, vector<vector<int>>& matrix) {
if (f[x][y] != -1) return f[x][y];
f[x][y] = 1;
for (int i = 0; i < 4; i ++) {
int a = x + dx[i], b = y + dy[i];
if (a >= 0 && a < n && b >= 0 && b < m && matrix[a][b] > matrix[x][y])
f[x][y] = max(f[x][y], dp(a, b, matrix) + 1);
}
return f[x][y];
}
};
