1 //1 使用邻接表 时间复杂度: O(n+e) 2 //递归 3 public void DFS(int v) 4 { 5 System.out.print(this.vexs[v].data + " "); 6 this.visited[v] = true; 7 8 for(ArcNode p = this.vexs[v].firstarc; p != null; p = p.nextarc) 9 if(this.visited[p.adjvex] == false) 10 this.DFS(p.adjvex); 11 } 12 13 //迭代(借助栈) 14 public void DFS(int v) 15 { 16 System.out.print(this.vexs[v].data + " "); 17 this.visited[v] = true; 18 Stack stack = new Stack(); 19 stack.push(vexs[v]); 20 21 for(int i = 0; i < this.vexs.length; i++) 22 this.vexs[i].next = this.vexs[i].firstarc; 23 24 while(!stack.isEmpty()) 25 { 26 VexNode q = stack.getTop(); 27 ArcNode p; 28 for(p = q.next; p != null && this.visited[p.adjvex] == true; p = p.nextarc); 29 if(p != null) 30 { 31 System.out.print(this.vexs[p.adjvex].data + " "); 32 this.visited[p.adjvex] = true; 33 stack.push(this.vexs[p.adjvex]); 34 q.next = p.nextarc; 35 } 36 else 37 stack.pop(); 38 } 39 }
1 //使用邻接矩阵 时间复杂度: O(n^2) 2 //递归 3 public void DFS(int v) 4 { 5 System.out.print(this.vexs[v] + " "); 6 this.visited[v] = true; 7 8 for(int i = 0; i < this.vexnum; i++) 9 if(this.adjacencyMatrix[v][i] == 1 && this.visited[i] == false) 10 this.DFS(i); 11 } 12 13 //迭代(借助栈) 14 public void DFS(int v) 15 { 16 System.out.print(this.vexs[v] + " "); 17 this.visited[v] = true; 18 Stack stack = new Stack(); 19 stack.push(v); 20 21 while(!stack.isEmpty()) 22 { 23 v = stack.getTop(); 24 int i; 25 for(i = 0; i < this.vexnum; i++) 26 if(this.adjacencyMatrix[v][i] == 1 && this.visited[i] == false) 27 { 28 System.out.print(this.vexs[i] + " "); 29 this.visited[i] = true; 30 stack.push(i); 31 break; 32 } 33 if(i == this.vexnum) 34 stack.pop(); 35 } 36 }
原文:https://www.cnblogs.com/Huayra/p/10816415.html