LeetCode 329 Longest Increasing Path in a Matrix (记忆化搜索)

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:

nums = [
  [9,9,4],
  [6,6,8],
  [2,1,1]
]

Return 4
The longest increasing path is [1, 2, 6, 9].

Example 2:

nums = [
  [3,4,5],
  [3,2,6],
  [2,2,1]
]

Return 4
The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed.

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

题目链接：https://leetcode.com/problems/longest-increasing-path-in-a-matrix/

题目分析：设dp[i][j]为到(i, j)这个位置时最长上升路径长度，击败了50%。。。

public class Solution {
    
    int [][]dp;
    int dx[] = {0, 0, 1, -1};
    int dy[] = {1, -1, 0, 0};
    
    int DFS (int x, int y, int n, int m, int[][] matrix) {
        if (dp[x][y] != 0) {
            return dp[x][y];
        }
        dp[x][y] = 1;
        for (int i = 0; i < 4; i ++) {
            int xx = x + dx[i];
            int yy = y + dy[i];
            if (xx >= 0 && xx < n && yy >= 0 && yy < m && matrix[xx][yy] < matrix[x][y]) {
                dp[x][y] = Math.max(dp[x][y], DFS(xx, yy, n, m, matrix) + 1);
            }   
        }
        return dp[x][y];
    }
    
    public int longestIncreasingPath(int[][] matrix) {
        int n = matrix.length;
        if (n == 0) {
            return 0;
        }
        int m = matrix[0].length;
        dp = new int[n + 1][m + 1];
        for (int i = 0; i < n; i ++) {
            Arrays.fill(dp[i], 0);
        }
        int ans = 1;
        for (int i = 0; i < n; i ++) {
            for (int j = 0; j < m; j ++) {
                ans = Math.max(ans, DFS(i, j, n, m, matrix));
            }
        }
        return ans;
    }
}