1、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.
2宵凌、問題描述：
- 在一個(gè)矩陣中找到一個(gè)最長的嚴(yán)格遞增的路徑郊供。要求不能對角線走。
3徐鹤、問題關(guān)鍵：
- DP垃环，能走到當(dāng)前點(diǎn)，說明比當(dāng)前點(diǎn)小返敬。
- 狀態(tài)變量：f[x][y]遂庄，表示走到（x， y）點(diǎn)的最長路徑劲赠。
- 狀態(tài)轉(zhuǎn)移：f[x][y] = max(f[x][y], dp(a, b) + 1); (a, b)是(x, y)周圍的點(diǎn)涛目。表示能夠從(a, b) 走到(x, y).
4、C++代碼：
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];
}
};
