[Swift]LeetCode1025. 除数博弈 | Divisor Game

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

Initially, there is a number N on the chalkboard.  On each player's turn, that player makes a move consisting of:

• Choosing any x with 0 < x < N and N % x == 0.
• Replacing the number N on the chalkboard with N - x.

Also, if a player cannot make a move, they lose the game.

Return True if and only if Alice wins the game, assuming both players play optimally.

Example 1:

Input: 2
Output: true Explanation: Alice chooses 1, and Bob has no more moves. 

Example 2:

Input: 3
Output: false Explanation: Alice chooses 1, Bob chooses 1, and Alice has no more moves.

Note:

1. 1 <= N <= 1000

---

爱丽丝和鲍勃一块儿玩游戏，他们轮流行动。爱丽丝先手开局。

最初，黑板上有一个数字 N 。在每一个玩家的回合，玩家须要执行如下操做：

• 选出任一 x，知足 0 < x < N 且 N % x == 0 。
• 用 N - x 替换黑板上的数字 N 。

若是玩家没法执行这些操做，就会输掉游戏。

只有在爱丽丝在游戏中取得胜利时才返回 True，不然返回 false。假设两个玩家都以最佳状态参与游戏。

示例 1：

输入：2
输出：true
解释：爱丽丝选择 1，鲍勃没法进行操做。

示例 2：

输入：3
输出：false
解释：爱丽丝选择 1，鲍勃也选择 1，而后爱丽丝没法进行操做。

提示：

1. 1 <= N <= 1000

---

Runtime: 4 ms

Memory Usage: 18.8 MB

1 class Solution {
2     func divisorGame(_ N: Int) -> Bool {
3         return N % 2 == 0        
4     }
5 }

---

Runtime: 4 ms

Memory Usage: 18.6 MB

 1 class Solution {
 2     func divisorGame(_ N: Int) -> Bool {
 3         var N = N
 4         var alice:Bool = false
 5         for x in 1..<N
 6         {
 7             alice.toggle()
 8             if N % x == 0
 9             {
10                 N -= x
11             }     
12         }
13         return alice
14     }
15 }

---

44ms

 1 class Solution {
 2     func divisorGame(_ N: Int) -> Bool {
 3         var dp = [Int:Bool]()
 4         dp[1] = false
 5         dp[2] = true
 6         
 7         var i = 3
 8         while i <= N{
 9             var x = i
10             var lose = false
11             dp[i] = false
12             repeat{
13                 x -= 1
14                 if (i%x == 0 && dp[i-x]! == false){
15                     dp[i] = true
16                 }
17             }while x > 1
18 
19             i += 1
20         }
21         
22         return dp[N]!
23     }
24 }