In the "100 game," two players take turns adding, to a running total, any integer from 1..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?

For example, two players might take turns drawing from a common pool of numbers of 1..15 without replacement until they reach a total >= 100.

Given an integer maxChoosableInteger and another integer desiredTotal, determine if the first player to move can force a win, assuming both players play optimally.

You can always assume that maxChoosableInteger will not be larger than 20 and desiredTotal will not be larger than 300.

Example

Input:maxChoosableInteger = 10desiredTotal = 11Output:falseExplanation:No matter which integer the first player choose, the first player will lose.The first player can choose an integer from 1 up to 10.If the first player choose 1, the second player can only choose integers from 2 up to 10.The second player will win by choosing 10 and get a total = 11, which is >= desiredTotal.Same with other integers chosen by the first player, the second player will always win.

这道题给了我们一堆数字，然后两个人，每人每次选一个数字，看数字总数谁先到给定值，有点像之前那道Nim Game，但是比那题难度大。我刚开始想肯定说用递归啊，结果写完发现TLE了，后来发现我们必须要优化效率，使用哈希表来记录已经计算过的结果。我们首先来看如果给定的数字范围大于等于目标值的话，直接返回true。如果给定的数字总和小于目标值的话，说明谁也没法赢，返回false。然后我们进入递归函数，首先我们查找当前情况是否在哈希表中存在，有的话直接返回即可。我们使用一个整型数按位来记录数组中的某个数字是否使用过，我们遍历所有数字，将该数字对应的mask算出来，如果其和used相与为0的话，说明该数字没有使用过，我们看如果此时的目标值小于等于当前数字，说明已经赢了，或者我们调用递归函数，如果返回false，说明也是第一个人赢了。为啥呢，因为当前我们已经选过数字了，此时就该对第二个人调用递归函数，只有他的结果是false，我们才能赢，所以此时我们标记true，返回true。如果遍历完所有数字，我们标记false，返回false，参见代码如下：

class Solution {public:    bool canIWin(int maxChoosableInteger, int desiredTotal) {        if (maxChoosableInteger >= desiredTotal) return true;        if (maxChoosableInteger * (maxChoosableInteger + 1) / 2 < desiredTotal) return false;        unordered_map<int, bool> m;        return canWin(maxChoosableInteger, desiredTotal, 0, m);    }    bool canWin(int length, int total, int used, unordered_map<int, bool>& m) {        if (m.count(used)) return m[used];        for (int i = 0; i < length; ++i) {            int cur = (1 << i);            if ((cur & used) == 0) {                if (total <= i + 1 || !canWin(length, total - (i + 1), cur | used, m)) {                    m[used] = true;                    return true;                }            }        }        m[used] = false;        return false;    }};

类似题目：

Nim Game  

参考资料：

https://discuss.leetcode.com/topic/68896/java-solution-using-hashmap-with-detailed-explanation/2

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