因为连续休息时间可能跨天,所以把两天拼起来,求 最长的连续1序列 就行了
时空复杂度
- 时间复杂度:$O(n)$
- 空间复杂度:$O(1)$
Java 代码
import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int T = scanner.nextInt();
while (T-- > 0) {
int n = scanner.nextInt();
// 这里数组长度为 2n,把两天拼起来
int[] nums = new int[n * 2];
for (int i = 0; i < n; i++) {
nums[i] = nums[i + n] = scanner.nextInt();
}
int count = 0, maxCount = 0;
for (int i = 0; i < n * 2; i++) {
if (nums[i] == 1) {
count++;
} else {
maxCount = Math.max(count, maxCount);
count = 0;
}
}
System.out.println(maxCount);
}
scanner.close();
}
}