商汤报数

第一题思路 二三题刚不动
package algorithm.problems;
import java.util.Scanner;
import org.junit.Test;
/**

• /
  class P{
  public int num = 0;
  public int state = 1;
  public P() {
  }
  public P(int num, int state) {

     this.num = num;
     this.state = state;

  }
  }
  public class YSF {
  public int[] ysf_1() {

     int[] idx = new int[2];
     Scanner sc = new Scanner(System.in);
     int n = sc.nextInt();
     int k = sc.nextInt();
     P[] arr = new P[n];
     for(int i = 0; i < n; ++i) {
         arr[i] = new P();
         (arr[i]).num = i + 1;
     }
     int CNT_1 = n;//剩多少个1
     int cnt = 0;
     int reset = 0;
     for(int j = 0; j < n; ++j) {
         if(reset == 1)
             cnt = 0;//重置
         reset = 0;
         if(arr[j].state == 1) {//表示在队列中
             ++cnt;
             if(cnt == 1) {//如果报1，就置0
                 arr[j].state = 0;
                 --CNT_1;
             }else if(cnt == k){
                 reset = 1;
             }
             if(CNT_1 == 2) {//如果最后剩余2个人了
                 break;//结束操作
             }
         }
         if(j == n - 1)
             j = -1;
         //下一次循环，可以用HashMap写，把报数为1的删除，可以减小复杂度，不贴了
     }
     int t = 0;
     for(int r = 0; r < n; ++r) {
         if(arr[r].state == 1) {
             idx[t] = arr[r].num;
             ++t;
             if(t == 2)
                 break;
         }
     }
     sc.close();
     return idx;

  }
  @Test
  public void test() {

     YSF Ysf = new YSF();
     int[] idx = Ysf.ysf_1();
     System.out.print(idx[0] + " " + idx[1]);

  }
  }