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• Multiple choice: 7 questions, 21 points
• Multiple select: 5 questions, 30 points
• Fill-in-the-blank: 2 questions, forgot how many points
• Short answer: 2 questions, don’t remember
• Programming: 3 problems at easy, easy, medium difficulty. Core code mode. Total code volume under 40 lines.

Finished the programming problems in 20 minutes, submitted, and went out to have fun. Tsk.

Short Answer 2 - Implementing a get Function

After the exam, I let Copilot auto-complete it. Hilarious.

let obj = { a: [{ b: { c: 1 } }] };
// console.log(obj.a[0].b.c)
function get(obj, str) {
  str = str.split('.');
  let len = str.length;
  for (let i = 0; i < len; ++i) {
    if (str[i] && str[i].indexOf('[') != -1) {
      let index = str[i].match(/\[(\d+)\]/)[1];
      let name = str[i].split('[')[0];
      if (name in obj) {
        obj = obj[name][index];
      } else {
        return undefined;
      }
    } else if (str[i] in obj && obj[str[i]]) {
      obj = obj[str[i]];
    } else {
      return undefined;
    }
  }
  return obj;
}
console.log(get(obj, 'a[0].b.c'));
console.log(get(obj, 'a[0].b.s'));
console.log(get(obj, 'a.s.q'));

Programming Problem 1 - Fibonacci

Fibonacci, one-liner: return n == 0 || n == 1? n: fib(n-1)+fib(n-2)

Programming Problem 2 - Bracket Matching

Solved instantly. Just use a stack, ignore * characters.

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param str string字符串 输入字符串
 * @return bool布尔型
 */
function isVerified(str) {
  // write code here
  let len = str.length;
  let s = [];
  for (let i = 0; i < len; ++i) {
    if (str[i] == '*') continue;
    if (str[i] == '{' || str[i] == '[' || str[i] == '(') {
      s.push(str[i]);
    } else {
      let top = s[s.length - 1];
      if ((str[i] == '}' && top == '{') || (str[i] == ']' && top == '[') || (str[i] == ')' && top == '(')) {
        s.pop();
      }
    }
  }
  return s.length == 0;
}
module.exports = {
  isVerified: isVerified,
};

Programming Problem 3 - Longest Increasing Subsequence Length

Done this so many times. Just DP.

• dp[i] represents the length of the longest increasing subsequence from index 0 to i. Each time, find the j from 0 to i-1 that can form an increasing sequence with the current i, and take the maximum as dp[i].

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 *
 * @param arr int整型一维数组 数组
 * @return int整型
 */
function lengthOfLis(arr) {
  // write code here
  let len = arr.length;
  let dp = new Array(len).fill(1);
  dp[0] = 1;
  for (let i = 1; i < len; ++i) {
    for (let j = 0; j < i; ++j) {
      if (arr[i] > arr[j]) dp[i] = Math.max(dp[j] + 1, dp[i]);
    }
  }
  return dp[len - 1];
}
module.exports = {
  lengthOfLis: lengthOfLis,
};
// 求数组最长递增子序列长度