}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 || ...
Can you solve this real interview question? Longest Increasing Path in a Matrix - Given an m x n integers matrix, return the length of the longest increasing path in matrix. From each cell, you can either move in four directions: left, right, up, or dow
best = 1 for dx, dy in dirs: nx, ny = x + dx, y + dy if -1 < nx < rows and -1 < ny < cols and matrix[nx][ny] > matrix[x][y]: best = max(best, dfs(nx, ny) + 1) return best for r in range(rows): for c in range(cols): ans = max(ans, dfs(r, c)) r...
}privateintdfs(int[][] matrix,intx,inty,int[][] record){if(record[x][y] >0)returnrecord[x][y]; record[x][y] =1;for(inti=0; i <4; i++) {intnx=x + xShift[i], ny = y + yShift[i];if(nx >=0&& nx < m && ny >=0&& ny < n && matrix[x][y] < matrix[nx][...
func longestIncreasingPath(matrix [][]int) int { cache, res := make([][]int, len(matrix)), 0 for i := 0; i < len(cache); i++ { cache[i] = make([]int, len(matrix[0])) } for i, v := range matrix { for j := range v { ...
[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 dp = [[0] * col for _ in range(row)] # 遍历每一个位置,以每一个位置为起点进行深度优先...
// 来源: http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-329-longest-increasing-path-in-a-matrix/ // Author: Huahua // Running time: 63 ms class Solution { public: int longestIncreasingPath(vector<vector<int>>& matrix) { if (matrix.empty()) return 0; int m = matrix.si...
publicstaticfinalint[][]dirs={{0,1},{1,0},{0,-1},{-1,0}};publicintlongestIncreasingPath(int[][]matrix){if(matrix.length==0)return0;intm=matrix.length,n=matrix[0].length;int[][]cache=newint[m][n];intmax=1;for(inti=0;i<m;i++){for(intj=0;j<n;j++){intlen=dfs(matri...
[]matrix,inti,intj,int[][]cache){if(cache[i][j]!=0)returncache[i][j];intmax=1;for(int[]dir:dirs){intx=i+dir[0];inty=j+dir[1];if(x<0||y<0||x>=m||y>=n||matrix[x][y]<=matrix[i][j])continue;intlen=bfs(matrix,x,y,cache);max=Math.max(max,1+len);}cache[i]...
zeros(0,StateSpace.NumStateVariables) (default) | real-valued M-by-N matrix NumStates— Number of state samples in path 0 (default) | nonnegative integer MaxNumStates— Maximum number of states allowed in path inf (default) | positive scalar integerObject...