猿辅导服务器端提前批二面(JAVA)
基础:
1.线程池
1.线程池
2.线程安全的方式
3. 写sql语句(太不自信了自己)
4.CMS
5.线程池设计
5.1 如何给URL爬虫设计线程池
5.2 如何给若干个计算设计线程池(计算之间不关联)
6.volatile实现
算法题:
1.给定一个不单调递增的链表,删除重复节点
1.给定一个不单调递增的链表,删除重复节点
2.给定一个二维矩阵m*n,有两个条件,找到等于target的最小坐标
2.1 data[i][j] <= data[i][j+1]
2.2 data[i+1][0] >= data[i][N-1]
public class Main { public static void main(String[] args) { int[][] data = new int[][]{{1,2,2,3},{3,4,5,5},{6,6,6,6},{7,8,9,9}}; int[] res = find(data,9); System.out.println(res[0]+":"+res[1]); } public static int[] find(int[][] data,int target){ int M = data.length; int N = data[0].length; int index = -1; for(int i = 0;i < M;i++){ if(data[i][0] <= target && target <= data[i][N-1]){ index = i; break; } } int left = 0,right = N-1; while(left <= right){ int mid = (left+right)/2; if(data[index][mid] == target){ while(mid>=0 && data[index][mid] == target) mid--; int[] res = new int[2]; res[0] = index; res[1] = mid+1; return res; }else if(data[index][mid] < target){ left = mid + 1; }else{ right = mid - 1; } } return null; } }
public static int[] find2(int[][] data,int target){ int M = data.length; int N = data[0].length; int ii = M-1, jj = 0; while(ii >= 0 && jj < N){ if(data[ii][jj] == target){ int x = ii, y = jj; while(ii >= 0 && data[ii][jj] == target){ x = ii; y = jj; ii = jj == 0 ? ii-1 : ii; jj = jj == 0 ? N-1 : jj-1; } int[] res = new int[2]; res[0] = x; res[1] = y; return res; }else if(data[ii][jj] < target){ jj++; }else{ ii--; } } return null; }以上是后来的两个思路。
第二题的时候,题目没看全,尴尬了,以后还是把语文学好先。
算法题还好,各位加油吧。还是为面试官打call吧,尽管我太菜。😃
#猿辅导##Java工程师#