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
题目难度：中等
题目大意：Alice和Bob轮流玩一个游戏，Alice先。一开始，堆里有n个石头，轮到某个人的时候，它可以移除堆里的平方数（就是1个，4个，9个...)个石头。如果轮到某人的时候，它不能移除石头，那么它就输了。
给定一个正整数n，如果Alice赢了，函数返回True，否则返回False。前提是两个人都能玩得很好。
思路：两个人都能玩的很好，都是最优策略，假定某个人面对n个石头的时候，F[n] = True表示赢, 否则为False.
当有n个石头的时候，轮到Alice移除，她可以选择移除$1^2$, $2^2$, ..., ${\lfloor \sqrt{n} \rfloor}^2$个，假如Alice 移除 $1^2$个石头，接下来剩下$n - 1$个石头，Bob来移除，如果
F[n - 1] = True, 表明，n个石头，Alice移除1个的话就输了，如果F[n - 1] = False，表明Alice赢了，这是一个子问题结构。
Alice在面对n个石头的时候，也可以选择移除4个，那么只要F[n - 4] = False, 那么Alice也能赢。
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] ....其中有一个为0，则F[n] = 1)
代码一：动态规划
C++代码
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代码：
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
时间复杂度:$O(n \sqrt{n})$, 空间复杂度：$O(n)$
思路二：记忆化的递归 （自己实现）
