In the "100 game" two players take turns adding, to a running total, any integer from 1 to 10. The player who first causes the running total to reach or exceed 100 wins.
What if we change the game so that players cannot re-use integers?
\n\nFor example, two players might take turns drawing from a common pool of numbers from 1 to 15 without replacement until they reach a total >= 100.
\n\nGiven two integers maxChoosableInteger and desiredTotal, return true if the first player to move can force a win, otherwise, return false. Assume both players play optimally.
\n
Example 1:
\n\n\nInput: maxChoosableInteger = 10, desiredTotal = 11\nOutput: false\nExplanation:\nNo matter which integer the first player choose, the first player will lose.\nThe first player can choose an integer from 1 up to 10.\nIf the first player choose 1, the second player can only choose integers from 2 up to 10.\nThe second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.\nSame with other integers chosen by the first player, the second player will always win.\n\n\n
Example 2:
\n\n\nInput: maxChoosableInteger = 10, desiredTotal = 0\nOutput: true\n\n\n
Example 3:
\n\n\nInput: maxChoosableInteger = 10, desiredTotal = 1\nOutput: true\n\n\n
\n
Constraints:
\n\n1 <= maxChoosableInteger <= 200 <= desiredTotal <= 300在 \"100 game\" 这个游戏中，两名玩家轮流选择从 1 到 10 的任意整数，累计整数和，先使得累计整数和 达到或超过 100 的玩家，即为胜者。
如果我们将游戏规则改为 “玩家 不能 重复使用整数” 呢？
\n\n例如，两个玩家可以轮流从公共整数池中抽取从 1 到 15 的整数（不放回），直到累计整数和 >= 100。
\n\n给定两个整数 maxChoosableInteger （整数池中可选择的最大数）和 desiredTotal（累计和），若先出手的玩家能稳赢则返回 true ，否则返回 false 。假设两位玩家游戏时都表现 最佳 。
\n\n
示例 1：
\n\n\n输入：maxChoosableInteger = 10, desiredTotal = 11\n输出：false\n解释：\n无论第一个玩家选择哪个整数，他都会失败。\n第一个玩家可以选择从 1 到 10 的整数。\n如果第一个玩家选择 1，那么第二个玩家只能选择从 2 到 10 的整数。\n第二个玩家可以通过选择整数 10（那么累积和为 11 >= desiredTotal），从而取得胜利.\n同样地，第一个玩家选择任意其他整数，第二个玩家都会赢。\n\n\n
示例 2:
\n\n\n输入：maxChoosableInteger = 10, desiredTotal = 0\n输出：true\n\n\n
示例 3:
\n\n\n输入：maxChoosableInteger = 10, desiredTotal = 1\n输出：true\n\n\n
\n\n
提示:
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