leetcode 464. Can I Win

464. Can I Win

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.

0、新学了一招。 bitset<20> mybit;

1、BFS + 记忆化搜索。记忆化搜索记忆的是 20bit组成的unsigned long这个数。

因为一共最多20个数可以用，如果用过，第i为就设为1.

2、之前自己没有用bitset，而是用vector<int>来记录，这样没法记忆化搜索。

class Solution {private:    unordered_map<unsigned long, bool> mymap;    int _maxInteger;    bool dp(bitset<20>& mybit, int remain)     {        if (mymap.find(mybit.to_ulong()) != mymap.end()) //如果之前出现过就可以直接用。记忆化搜索。            return mymap[mybit.to_ulong()];        if(remain <= 0)         {            mymap[mybit.to_ulong()] = false;            return false;        }        for(int i = 1; i <= _maxInteger; i++)          {            if(!mybit[i-1])             {                mybit.set(i-1);//这一位置为1                bool temp = dp(mybit, remain - i);                mybit.reset(i-1); //这一位清为0                                if(!temp)                 {                    mymap[mybit.to_ulong()] = true; //记忆化搜索。在返回之前保存这一次的结果。以后可以直接用。                    return true;                }            }        }        mymap[mybit.to_ulong()] = false;        return false;    }public:    bool canIWin(int maxChoosableInteger, int desiredTotal)     {        _maxInteger = maxChoosableInteger;        bitset<20> mybit;        mybit.reset(); //所有位置为0        if (desiredTotal <= 0) return true;        if ((1+maxChoosableInteger) * maxChoosableInteger / 2 < desiredTotal) return false;             return dp(mybit, desiredTotal);    }};