time. Find the cycle created by the edge, and contract the entire cycle into one vertex. Repeat for the entire Dijkstra run. This should take the same total time complexity with Dijkstra. The edges in the short
0133-clone-graph.py 0134-gas-station.py 0136-single-number.py 0138-copy-list-with-random-pointer.py 0139-word-break.py 0141-linked-list-cycle.py 0143-reorder-list.py 0146-lru-cache.py 0150-evaluate-reverse-polish-notation.py 0152-maximum-product-subarray.py 0153-fin...
else { parent[y] = x; // 父节点 x // 并查集检查环 if (!uf.union(x, y)) cycle = i; } } if (conflict < 0) return edges[cycle]; else { int[] conflictEdge = edges[conflict]; if (cycle >= 0) conflictEdge[0] = parent[conflictEdge[1]]; return conflictEdge; } } } 1. ...