LeetCode-329. Longest Increasing Path in a Matrix

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.

题解：

第一次对每个点进行dfs，发现超时了

class Solution {
public:
  void dfs(int n, int m, vector<vector<int>> matrix, int &ans, int d, int i, int j) {
     
    if (i + 1 >= 0 && i + 1 < n && j >= 0 && j < m && matrix[i + 1][j] > matrix[i][j]) {
      dfs(n, m, matrix, ans, d + 1, i + 1, j);
    }
    if (i >= 0 && i < n && j + 1 >= 0 && j + 1 < m && matrix[i][j + 1] > matrix[i][j]) {
      dfs(n, m, matrix, ans, d + 1, i, j + 1);
    }
    if (i - 1 >= 0 && i - 1 < n && j >= 0 && j < m && matrix[i - 1][j] > matrix[i][j]) {
      dfs(n, m, matrix, ans, d + 1, i - 1, j);
    }
    if (i >= 0 && i < n && j - 1 >= 0 && j - 1 < m && matrix[i][j - 1] > matrix[i][j]) {
      dfs(n, m, matrix, ans, d + 1, i, j - 1);
    }
      ans = max(ans, d);
  }
  int longestIncreasingPath(vector<vector<int>>& matrix) {
    int n = matrix.size();
    if (matrix.empty() == true) {
      return 0;
    }
    int m = matrix[0].size();
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
          dfs(n, m, matrix, ans, 1, i, j);
      }
    }
    return ans;
  }
};

看讨论区是使用记忆化搜索，学习了一下

class Solution {
public:
    int find(vector<vector<int>>& matrix, vector<vector<int>>& dp, int n, int m, int x, int y) {
        if (x >= n || x < 0 || y >= m || y < 0) {
            return 0;
        }
        if (dp[x][y] != 0) {
            return dp[x][y];
        }
        int up = 0, down = 0, left = 0, right = 0;
        if (y - 1 >= 0 && matrix[x][y - 1] < matrix[x][y]) {
            up = find(matrix, dp, n, m, x, y - 1);
        }
        if (y + 1 < m && matrix[x][y + 1] < matrix[x][y]) {
            down = find(matrix, dp, n, m, x, y + 1);
        }
        if (x - 1 >= 0 && matrix[x - 1][y] < matrix[x][y]) {
            left = find(matrix, dp, n, m, x - 1, y);
        }
        if (x + 1 < n && matrix[x + 1][y] < matrix[x][y]) {
            right = find(matrix, dp, n, m, x + 1, y);
        }
        dp[x][y] = 1 + max({up, down, left, right});
        return dp[x][y];
    }
    int longestIncreasingPath(vector<vector<int>>& matrix) {
        if (matrix.empty() == true) {
            return 0;
        }
        int n = matrix.size(), m = matrix[0].size(), res = 0;
        vector<vector<int>> dp(n, vector<int>(m, 0));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                res = max(res, find(matrix, dp, n, m, i, j));
            }
        }
        return res;
    }
};