classSolution {public: vector<vector<int>> dirs = {{0, -1}, {-1,0}, {0,1}, {1,0}};intlongestIncreasingPath(vector<vector<int>>&matrix) {if(matrix.empty() || matrix[0].empty())return0;intres =1, m = matrix.size(), n = matrix[0].size(); vector<vector<int>> dp(m, ...
代码如下: classSolution{public:intlongestIncreasingPath(vector<vector<int>>&matrix){introws=matrix.size();if(rows==0)return0;intcols=matrix[0].size();vector<int>all;intdirs[4][2]={{0,1},{0,-1},{1,0},{-1,0}};unordered_map<int,vector<pair<int,int>>>maps;for(inti=0;i<rows...
第一种方法,递归。很明显,时间超时,通不过。 1classSolution {2publicintlongestIncreasingPath(int[][] matrix) {3if(matrix ==null|| matrix.length ==0|| matrix[0].length ==0)return0;4intmax =0;5boolean[][] visited =newboolean[matrix.length][matrix[0].length];6for(inti =0; i < matr...
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...
class Solution: 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 ...
// Author: Huahua // Running time: 36 ms class Solution { public: int longestIncreasingPath(vector<vector<int>>& matrix) { if (matrix.empty()) return 0; m_ = matrix.size(); n_ = matrix[0].size(); dp_ = vector<vector<int>>(m_, vector<int>(n_, -1)); int ans = 0; fo...
这个是在退出的时候标记visited.因为是严格递增的,所以不需要避免环。 所以不需要visiting, 只需要一个visited. classSolution{int[][]OFFSETS;publicintlongestIncreasingPath(int[][]matrix){if(matrix==null||matrix.length==0||matrix[0].length==0)return0;int[]ans=newint[1];OFFSETS=newint[][]{{-...
class Solution: def longestIncreasingPath(self, matrix: List[List[int]]) -> int: def dfs(i, j): if not dp[i][j]: val = matrix[i][j] dp[i][j] = 1 + max( dfs(i - 1, j) if i and val > matrix[i - 1][j] else 0, ...
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 { ...
Explanation: The longest increasing path is [3, 4, 5, 6]. Moving diagonally is not allowed. 1. 2. 3. 4. 5. 6. 7. 8. 题解: 第一次对每个点进行dfs,发现超时了 classSolution{ public: voiddfs(intn,intm,vector<vector<int>>matrix,int&ans,intd,inti,intj) { ...