标签:div sem bool href empty lin click bfs move
bfs可以实现。这里注意的是,机器人从00出发,我们只要向右,向下走就可以了
public static int movingCount(int m, int n, int k) { int[] dx=new int[] {-1,1,0,0}; int[] dy = new int[]{0,0,-1,1}; boolean[][] vis = new boolean[m][n]; Queue<int[]> q = new LinkedList<>(); q.offer(new int[]{0,0}); vis[0][0]=true; int ans = 1; while (!q.isEmpty()){ int[] nowN = q.poll(); for(int i=0;i<4;i++){ int nextX = nowN[0]+dx[i]; int nextY = nowN[1]+dy[i]; //boolean flag = canMove(nextX,nextY,k); if(nextX>=0&&nextX<m &&nextY>=0&&nextY<n&&!vis[nextX][nextY]&&canMove(nextX,nextY,k)){ ans++; vis[nextX][nextY]=true; q.offer(new int[]{nextX,nextY}); } } } return ans; } private static boolean canMove(int nextX, int nextY,int k) { int sum=0; while (nextX>0){ sum+=nextX%10; nextX/=10; } while (nextY>0){ sum+=nextY%10; nextY/=10; } return sum<=k ? true:false; }
标签:div sem bool href empty lin click bfs move
原文地址:https://www.cnblogs.com/superxuezhazha/p/12662575.html