题目：
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 = 10 desiredTotal = 11 Output: false Explanation: 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.
代码：
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;
}
};
思路：
如果maxchoosableInteger>=desiredInteger，那么返回true，第一个人可以赢;如果maxchoosable*(maxchoosable+1)/2<desiredInteger,那么返回false,谁也赢不了。下面调用一个递归，参数total表示desiredtotal,used记录i有没有被用过，使用位运算进行记录，维持一个字典，这个字典表示给定的used是否能用，返回记录。