记录笔试0829

第一部分：单项选择

第二部分：多项选择

第三部分：编程三道题

第一题：SQL

有一个视频发布记录表，有视频主键，发布者id，发布视频时长，请在这个表中，请用MYSQL查询，在视频数据表中，筛选发布视频平均时长大雨300秒的up主，按照up主全部视频的平均时长倒序，视频ID升序，返回第三行到第六行的视频id

SELECT video_id 
FROM video_table
WHERE uploader_id IN (
  SELECT uploader_id
  FROM video_table 
  GROUP BY uploader_id
  HAVING AVG(video_duration) > 300
) 
ORDER BY (
  SELECT AVG(video_duration) 
  FROM video_table AS t 
  WHERE t.uploader_id = video_table.uploader_id
) DESC, video_id ASC
LIMIT 4 OFFSET 2;

第二题：动态规划

给定两个字符串，返回使两个字符串相等所需要删除字符的ASCII值的最小和

public class Main {
  public static int minimumDeleteSum(String s1, String s2) {
    int[][] dp = new int[s1.length() + 1][s2.length() + 1];
    for (int i = s1.length() - 1; i >= 0; i--) {
      dp[i][s2.length()] = dp[i + 1][s2.length()] + s1.codePointAt(i);
    }
    for (int j = s2.length() - 1; j >= 0; j--) {
      dp[s1.length()][j] = dp[s1.length()][j + 1] + s2.codePointAt(j);
    }
    for (int i = s1.length() - 1; i >= 0; i--) {
      for (int j = s2.length() - 1; j >= 0; j--) {
        if (s1.charAt(i) == s2.charAt(j)) {
          dp[i][j] = dp[i + 1][j + 1];
        } else {
          dp[i][j] = Math.min(dp[i + 1][j] + s1.codePointAt(i), dp[i][j + 1] + s2.codePointAt(j));
        }
      }
    }
    return dp[0][0];
  }

  public static void main(String[] args) {
    String s1 = "delete";
    String s2 = "leet";
    System.out.println(minimumDeleteSum(s1, s2));
  }
}

第三题：二叉树

给定一个二叉树的root，返回最长路径的长度，这个路径中每个节点具有相同的值，这条路径可以经过也可以不经过根节点，两个节点之间的长度由他们之间的边数表示

public int longestUnivaluePath(TreeNode root) {
  int[] maxLen = new int[1];
  maxLen[0] = 0;
  longestPath(root, maxLen);
  return maxLen[0];
}

private int longestPath(TreeNode node, int[] maxLen) {
  if (node == null) {
  return 0; 
  }
  
  int left = longestPath(node.left, maxLen);
  int right = longestPath(node.right, maxLen);
  
  int leftLen = 0, rightLen = 0;
  if (node.left != null && node.left.val == node.val) {
  leftLen = left + 1;
  }
  if (node.right != null && node.right.val == node.val) {
  rightLen = right + 1;
  }
  
  maxLen[0] = Math.max(maxLen[0], leftLen + rightLen);
  return Math.max(leftLen, rightLen);
}

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