【leetcode】329. Longest Increasin

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];
    }
};