847. 访问所有节点的最短路径 - 存在一个由 n 个节点组成的无向连通图,图中的节点按从 0 到 n - 1 编号。 给你一个数组 graph 表示这个图。其中,graph[i] 是一个列表,由所有与节点 i 直接相连的节点组成。 返回能够访问所有节点的最短路径的长度。你可以在任一节点开始
设计最短路径 用bfs 天然带最短路径 每一个状态是 当前的阶段 和已经访问过的节点 下面是正确但是超时的代码 classSolution:defshortestPathLength(self, graph):""" :type graph: List[List[int]] :rtype: int """N=len(graph) Q=collections.deque([(1<< x, x)forxinrange(N)]) D=collections.de...
Github 同步地址: https://github.com/grandyang/leetcode/issues/864 参考资料: https://leetcode.com/problems/shortest-path-to-get-all-keys/ https://leetcode.com/problems/shortest-path-to-get-all-keys/discuss/146878/Java-BFS-Solution https://leetcode.com/problems/shortest-path-to-get-all-keys...
pair<int, int> BFS(const vector<string>& grid, int x, int y, string keysSeached, char curKey, int& len) { if(x < 0 || y < 0 || x >= grid.size() || y >= grid[x].size()) { return {-1, -1}; } // 标记走过的路径 vector<string> gridFlag(grid); gridFlag[x][y...
int shortestPathLength(vector<vector<int>>& graph) { const int n = graph.size(); const int kAns = (1 << n) - 1; queue<pair<int, int>> q; vector<vector<int>> visited(n, vector<int>(1 << n)); for (int i = 0; i < n; ++i) q.push({i, 1 << i}); int steps =...
looks like leetcode copied yourproblem →Reply nguyenquocthao00 13 months ago,#| ←Rev.3→0 My solution: 1. Find all the edges that are a part ofanyshortest path 2. Find a shortest path and save to an array 3. From all the other edges that are not part of the shortest path from...
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leetcode 864. Shortest Path to Get All Keys 说一下简单思路 因为钥匙数目很少,可以直接用二进制保存所得的钥匙状态,所以,用6把钥匙+现在所在地点+已走的路径的长度来表示不同的状态,用队列queue实现广搜算法即可 ps:我之前用新建一个类来保存6把钥匙+当前所在地点,表示已经走过的状态,发现在重载<运算符(即...
(n,false));unordered_map<char,int>distances;queue<pair<int,int>>q;q.push({r,c});visited[r][c]=true;while(!q.empty()){for(inti=q.size();i>0;i--){r=q.front().first;c=q.front().second;q.pop();if(grid[r][c]!=source and grid[r][c]!='.'){distances[grid[r][c]...