最长连续不重复子序列
- 双指针[j, i]维护的是以a[i]结尾的最长连续不重复子序列
- [j, i - 1]是前一步得到的最长连续不重复子序列,所以如果[j, i]中有重复元素,一定是a[i],所以[j, i - 1]中一定有一个数字与a[i]重复,因此右移j直到a[i]不重复(通过次数来判断)为止
- 数组s记录子序列a[j ~ i]中各元素出现次数
i移动到第二个9,次数为2,j减去1并且右移1位到3
i移动到第二个3,次数为2,j减去1并且右移1位到6
每次都是先减去1再右移,保证每个数字不重复
public class Main {public static void main(String[] args) {int[] a = new int[1000001];int[] s = new int[1000001];Scanner in = new Scanner(System.in);int n = in.nextInt();for(int i = 0; i < n; i ++)a[i] = in.nextInt();int res = 0;for(int i = 0, j = 0; i < n; i ++) {s[a[i]] ++;while(s[a[i]] > 1) {s[a[j]] --;j ++;}res = Math.max(res, i - j + 1);}System.out.println(res);}
}
数组元素的目标和
a数组选的值越大,那么b数组选的值越小,x是固定的
public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();int m = in.nextInt();int x = in.nextInt();int[] a = new int[n + 10];int[] b = new int[m + 10];for(int i = 0; i < n; i ++)a[i] = in.nextInt();for(int i = 0; i < m; i ++)b[i] = in.nextInt();for(int i = 0, j = m - 1; i < n; i ++) {while(a[i] + b[j] > x && j >= 0)j --;if(a[i] + b[j] == x && j >= 0) {System.out.println(i + " " + j);break;}}}
}
判断子序列
只需要找到一种情况即可
j 完全遍历一遍,找到的话 i 就往后移动一位,如果 i == n 说明全部找到
public class Main {public static void main(String[] args) {Scanner in = new Scanner(System.in);int n = in.nextInt();int m = in.nextInt();int[] a = new int[n + 10];int[] b = new int[m + 10];for(int i = 0; i < n; i ++)a[i] = in.nextInt();for(int i = 0; i < m; i ++)b[i] = in.nextInt();int i = 0, j = 0;while(i < n && j < m) {if(a[i] == b[j])i ++;j ++;}if(i == n)System.out.println("Yes");elseSystem.out.println("No");}
}