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

石子游戏IV。

Alice 和 Bob 两个人轮流玩一个游戏，Alice 先手。

一开始，有 n 个石子堆在一起。每个人轮流操作，正在操作的玩家可以从石子堆里拿走 任意 非零 平方数 个石子。

如果石子堆里没有石子了，则无法操作的玩家输掉游戏。

给你正整数 n ，且已知两个人都采取最优策略。如果 Alice 会赢得比赛，那么返回 True ，否则返回 False 。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/stone-game-iv
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

题意不难理解，每轮游戏，每个玩家需要拿走一个非零的平方数。所以我们需要从1到N开始，对每一种情况的石子数量做判断，看他的DP值是true还是false。这里我们创建一个长度为N + 1的boolean数组，dp的含义是当数字为 i 的时候，到底是谁赢。这里我们定义dp[i] = true是Alice赢，反之则是Bob赢。

从1到N判断DP值，并且试着从N中拿出一个平方数，如果能使得剩下的数字的DP值为false，那么Alice就能赢。

时间O(n * 根号n)

空间O(n)

Java实现

 1 class Solution {
 2     public boolean winnerSquareGame(int n) {
 3         boolean[] dp = new boolean[n + 1];
 4         for (int i = 1; i <= n; i++) {
 5             for (int k = 1; k * k <= i; k++) {
 6                 if (!dp[i - k * k]) {
 7                     dp[i] = true;
 8                     break;
 9                 }
10             }
11         }
12         return dp[n];
13     }
14 }

 

LeetCode 题目总结