leetcode 1510. Stone Game IV

Alice and Bob take turns playing a game, with Alice starting first.

Initially, there are n stones in a pile.  On each player's turn, that player makes a move consisting of removing any non-zero square number of stones in the pile.

Also, if a player cannot make a move, he/she loses the game.

Given a positive integer n. Return True if and only if Alice wins the game otherwise return False, assuming both players play optimally.

 

Example 1:

Input: n = 1
Output: true
Explanation: Alice can remove 1 stone winning the game because Bob doesn't have any moves.

Example 2:

Input: n = 2
Output: false
Explanation: Alice can only remove 1 stone, after that Bob removes the last one winning the game (2 -> 1 -> 0).

Example 3:

Input: n = 4
Output: true
Explanation: n is already a perfect square, Alice can win with one move, removing 4 stones (4 -> 0).

Example 4:

Input: n = 7
Output: false
Explanation: Alice can't win the game if Bob plays optimally.
If Alice starts removing 4 stones, Bob will remove 1 stone then Alice should remove only 1 stone and finally Bob removes the last one (7 -> 3 -> 2 -> 1 -> 0). 
If Alice starts removing 1 stone, Bob will remove 4 stones then Alice only can remove 1 stone and finally Bob removes the last one (7 -> 6 -> 2 -> 1 -> 0).

Example 5:

Input: n = 17
Output: false
Explanation: Alice can't win the game if Bob plays optimally.

 

Constraints:

• 1 <= n <= 10^5

 

題目難度：中等

題目大意：Alice和Bob輪流玩一個遊戲，Alice先。一開始，堆裏有n個石頭，輪到某個人的時候，它可以移除堆裏的平方數（就是1個，4個，9個...)個石頭。如果輪到某人的時候，它不能移除石頭，那麼它就輸了。

給定一個正整數n，如果Alice贏了，函數返回True，否則返回False。前提是兩個人都能玩得很好。

 

思路：兩個人都能玩的很好，都是最優策略，假定某個人面對n個石頭的時候，F[n] = True表示贏, 否則爲False.

當有n個石頭的時候，輪到Alice移除，她可以選擇移除$1^2$, $2^2$, ..., ${\lfloor \sqrt{n} \rfloor}^2$個，假如Alice 移除 $1^2$個石頭，接下來剩下$n - 1$個石頭，Bob來移除，如果

F[n - 1] = True, 表明，n個石頭，Alice移除1個的話就輸了，如果F[n - 1] = False，表明Alice贏了，這是一個子問題結構。

Alice在面對n個石頭的時候，也可以選擇移除4個，那麼只要F[n - 4] = False, 那麼Alice也能贏。

F[n] = (1 - F[n - 1]) | (1 - F[n - 4]) | ... | (1 - F[n - ${\lfloor \sqrt{n} \rfloor}^2$] (或F[n - 1],F[n -4] ....其中有一個爲0，則F[n] = 1)

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代碼一：動態規劃

C++代碼

class Solution {
public:
    bool winnerSquareGame(int n) {
        vector<bool> f(n + 1, 0);
        for (int i = 1; i < n + 1; ++i) {
            // f[i] = 0;
            for (int j = 1; j * j <= i; ++j) {
                if (f[i - j * j] == false) {
                    f[i] = true;
                    break;
                }
                // f[i] = f[i] | (1 - f[i - j * j]);
                // if (f[i] == 1)
                //     break;
            }
        }
        return f[n];
    }
};

python3代碼：

 1 class Solution:
 2     def winnerSquareGame(self, n: int) -> bool:
 3         dp = [False] * (n + 1)
 4         for i in range(1, n + 1):
 5             j = 1
 6             while j * j <= i:
 7                 if dp[i - j * j] == False:
 8                     dp[i] = True
 9                     break
10                 j += 1
11         return dp[n]
12             

時間複雜度:$O(n \sqrt{n})$, 空間複雜度：$O(n)$

思路二：記憶化的遞歸 （自己實現）