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 = 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.

public class Solution {
    Map<Integer, Boolean> map;
    boolean[] used;
    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        int sum = (1+maxChoosableInteger)*maxChoosableInteger/2;
        if(sum<desiredTotal) return false;
        if(desiredTotal<=0) return true;
        map = new HashMap();
        used= new boolean[maxChoosableInteger+1];
        return dp(desiredTotal);
    }
    private boolean dp(int total){
        if(total<=0) return false;
        int key = format(used);
        if(!map.containsKey(key)){
            for(int i=1;i<used.length;i++){
                if(!used[i]){
                    used[i]=true;
                    if(!dp(total-i)){
                        map.put(key,true);
                        used[i]=false;
                        return true;
                    }
                    used[i]=false;
                }
            }
            map.put(key,false);
        }
        return map.get(key);
    }
    private int format(boolean[] used){
        int num =0;
        for(boolean b:used){
            num<<=1;
            if(b) num|=1;
        }
        return num;
    }
}

总结：不可以用boolean数组，因为函数浅拷贝。 Map<boolean[], Boolean>  ?  Obviously we cannot , because the if we use boolean[] as a key, the reference to boolean[] won't reveal the actual content in boolean[]. reference 173 ms

Most of "game playing" questions can be solved using the top-down DP approach, which "brute-forcely" simulates every possible state of the game. The key part for the top-down dp strategy is that we need to avoid repeatedly solving sub-problems. Instead, we should use some strategy to "remember" the outcome of sub-problems. Then when we see them again, we instantly know their result. 

public class Solution {
    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
        if (desiredTotal<=0) return true;
        if (maxChoosableInteger*(maxChoosableInteger+1)/2<desiredTotal) return false;
        return canIWin(desiredTotal, new int[maxChoosableInteger], new HashMap<>());
    }
    private boolean canIWin(int total, int[] state, HashMap<String, Boolean> hashMap) {
        String curr=Arrays.toString(state);
        if (hashMap.containsKey(curr)) return hashMap.get(curr);
        for (int i=0;i<state.length;i++) {
            if (state[i]==0) {
                state[i]=1;
                if (total<=i+1 || !canIWin(total-(i+1), state, hashMap)) {
                    hashMap.put(curr, true);
                    state[i]=0;
                    return true;
                }
                state[i]=0;
            }
        }
        hashMap.put(curr, false);
        return false;
    }
}

745ms，调用函数反而慢了。