375. Guess Number Higher or Lower II

375. Guess Number Higher or Lower IIopen in new window

🟠 🔖 数学 动态规划 博弈 🔗 LeetCodeopen in new window

题目

We are playing the Guessing Game. The game will work as follows:

I pick a number between 1 and n.
You guess a number.
If you guess the right number, you win the game.
If you guess the wrong number, then I will tell you whether the number I picked is higher or lower , and you will continue guessing.
Every time you guess a wrong number x, you will pay x dollars. If you run out of money, you lose the game.

Given a particular n, return the minimum amount of money you need to guarantee a win regardless of what number I pick.

Example 1:

Input: n = 10
Output: 16
Explanation: The winning strategy is as follows:
The range is [1,10]. Guess 7.
If this is my number, your total is $0. Otherwise, you pay $7.
If my number is higher, the range is [8,10]. Guess 9.
If this is my number, your total is $7. Otherwise, you pay $9.
If my number is higher, it must be 10. Guess 10. Your total is $7 + $9 = $16.
If my number is lower, it must be 8. Guess 8. Your total is $7 + $9 = $16.
If my number is lower, the range is [1,6]. Guess 3.
If this is my number, your total is $7. Otherwise, you pay $3.
If my number is higher, the range is [4,6]. Guess 5.
If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $5.
If my number is higher, it must be 6. Guess 6. Your total is $7 + $3 + $5 = $15.
If my number is lower, it must be 4. Guess 4. Your total is $7 + $3 + $5 = $15.
If my number is lower, the range is [1,2]. Guess 1.
If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $1.
If my number is higher, it must be 2. Guess 2. Your total is $7 + $3 + $1 = $11.
The worst case in all these scenarios is that you pay $16. Hence, you only need $16 to guarantee a win.

Example 2:

Input: n = 1
Output: 0
Explanation: There is only one possible number, so you can guess 1 and not have to pay anything.

Example 3:

Input: n = 2
Output: 1
Explanation: There are two possible numbers, 1 and 2.
Guess 1.
If this is my number, your total is $0. Otherwise, you pay $1.
If my number is higher, it must be 2. Guess 2. Your total is $1.
The worst case is that you pay $1.

Constraints:

1 <= n <= 200

题目大意

我们正在玩一个猜数游戏，游戏规则如下：

我从 1 到 n 之间选择一个数字。
你来猜我选了哪个数字。
如果你猜到正确的数字，就会 赢得游戏 。
如果你猜错了，那么我会告诉你，我选的数字比你的 更大或者更小 ，并且你需要继续猜数。
每当你猜了数字 x 并且猜错了的时候，你需要支付金额为 x 的现金。如果你花光了钱，就会输掉游戏。

给你一个特定的数字 n ，返回能够 确保你获胜 的最小现金数，不管我选择那个数字 。

解题思路

用 f(i, j) 表示在范围 [i, j] 内确保胜利的最少金额，目标是计算 f(1, n)。

假设第一次猜的数字是 x 并且猜错，则需要支付金额 x，当 x 大于所选数字时，为了确保胜利还需要支付的金额是 f(1, x - 1)，当 x 小于所选数字时，为了确保胜利还需要支付的金额是 f(x + 1, n)。为了在任何情况下都能确保胜利，应考虑最坏情况，计算 f(1, n) 时应取上述两者的最大值：f(1, n) = x + max(f(1, x − 1), f(x + 1, n))。

由于 f(1, x - 1) 和 f(x + 1, n) 都是比原始问题 f(1, n) 规模更小的问题，因此可以使用动态规划的方法求解。

为了将确保胜利的金额最小化，需要遍历从 1 到 n 的所有可能的 x，使得 f(1, n) 的值最小：

f(1, n) = min{x + max⁡(f(1, x − 1), f(x + 1, n))} (⁡1 ≤ x ≤ n)

创建行数和列数都是 n + 1 的二维数组 f，其中 f[i][j] 即为状态 f(i, j)。

在根据状态转移方程计算时需要注意下标的边界问题，当 j = n 时，如果 x = j 则 x + 1 > n，此时 f[x][j] 会出现下标越界。为了避免出现下标越界，计算 f[i][j] 的方法是：首先令 f[i][j] = j + f[i][j - 1]，然后遍历 i ≤ x < j 的每个 x，更新 f[i][j] 的值。

代码

/**
 * @param {number} n
 * @return {number}
 */
var getMoneyAmount = function (n) {
  const f = new Array(n + 1).fill(0).map(() => new Array(n + 1).fill(0));
  for (let i = n - 1; i >= 1; i--) {
    for (let j = i + 1; j <= n; j++) {
      f[i][j] = j + f[i][j - 1];
      for (let x = i; x < j; x++) {
        f[i][j] = Math.min(f[i][j], x + Math.max(f[i][x - 1], f[x + 1][j]));
      }
    }
  }
  return f[1][n];
};

375. Guess Number Higher or Lower II

# 375. Guess Number Higher or Lower IIopen in new window

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375. Guess Number Higher or Lower IIopen in new window

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