}intdfs(vector<vector<int>> &matrix, vector<vector<int>> &dp,inti,intj) {if(dp[i][j])returndp[i][j];intmx =1, m = matrix.size(), n = matrix[0].size();for(auto a : dirs) {intx = i + a[0], y = j + a[1];if(x <0|| x >= m || y <0|| y >= n || ...
代码如下: classSolution{public:intlongestIncreasingPath(vector<vector<int>>&matrix){introws=matrix.size();if(rows==0)return0;intcols=matrix[0].size();intlongest=0;vector<vector<int>>cache(rows,vector<int>(cols,0));vector<vector<int>>dirs={{0,1},{0,-1},{1,0},{-1,0}};for(int...
if not matrix: return 0 rows, cols, ans = len(matrix), len(matrix[0]), 0 dirs = [(0, 1), (0, -1), (1, 0), (-1, 0)] @lru_cache(None) def dfs(x, y): best = 1 for dx, dy in dirs: nx, ny = x + dx, y + dy if -1 < nx < rows and -1 < ny < cols...
int lens[];// 记住每个节点作为起点的最大长度 public int longestIncreasingPath(int[][] matrix) { if (matrix.length == 0) return 0; int rows = matrix.length, cols = matrix[0].length; int[] digree = new int[rows * cols];// 入度 lens = new int[rows * cols]; // 初始化邻接表...
classSolution {publicintlongestIncreasingPath(int[][] matrix) {intans = 0;if(matrix==null|| matrix.length==0)return0;intmemo[][] =newint[matrix.length][matrix[0].length];for(inti=0;i<matrix.length;i++){for(intj=0;j<matrix[0].length;j++){ ...
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). ...
def longestIncreasingPath(self, matrix: [[int]]) -> int: # 如果矩阵为空,返回 0 if not matrix or not matrix[0]: return 0 # 获取矩阵的行数和列数 row, col = len(matrix), len(matrix[0]) # 记忆化递归,记录每个位置的最大值
Leetcode 329. Longest Increasing Path in a Matrix 来自2050 . 目录 收起 分析 dfs + 记忆化(源自huahua) dp代码(源自huahua) 分析 引用huahua酱的一幅图 dfs + 记忆化 和dp 都是将状态从高层底层 引用huahua的一个评论:"dp的核心思想是从规模较小的解就构建规模较大的解 以较大的数出发的最长路径...
classSolution{public:vector<vector<int>>vis;int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};int n,m;intlongestIncreasingPath(vector<vector<int>>&matrix){n=matrix.size();if(n==0)return0;m=matrix[0].size();vis=matrix;for(int i=0;i<n;i++){for(int j=0;j<m;j++){vis...
Longest Increasing Path in a Matrix 题目链接:https://leetcode.com/problems... dfs + 记忆化搜索,用一个二维dp记录找到的最长路径的长度,如果发现dpi != 0,证明这个点被找过,不用重复。Number of Islands和这题一个思路。 public class Solution { ...