题解 | #交叉线#
交叉线
https://www.nowcoder.com/practice/54fe00d9b0e14688bd3d31ad539b929c
代码实现：
import java.util.*;
public class Main{
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
while (T-- > 0){
int n = in.nextInt();//端点数
int[] arr = new int[n];
for (int i = 0; i < n; i++) {//输入端点
arr[i] = in.nextInt();
}
List<int[]> cir = new ArrayList<>();
for (int i = 0; i < n - 1; i++) {//储存圆
int left = Math.min(arr[i],arr[i + 1]);//左端点
int right = Math.max(arr[i],arr[i + 1]);//右端点
cir.add(new int[]{left,right});
}
//题意为所有的圆都不存在相交，才返回false
boolean flag = false;
for (int i = 1; i < cir.size(); i++) {
for (int j = 0; j < i; j++) {
//先设置两个圆的端点
int left1 = cir.get(i)[0],right1 = cir.get(i)[1];//圆1的端点
int left2 = cir.get(j)[0],right2 = cir.get(j)[1];//圆2的端点
//两圆相交的情况
if (left1 < left2 && right1 < right2 && right1 > left2){
flag = true;
break;
}else if (left1 > left2 && left1 < right2 && right1 > right2){
flag = true;
break;
}
}
}
if (flag){
System.out.println("y");
}else {
System.out.println("n");
}
}
}
}
