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.


這個還沒看
https://www.youtube.com/watch?v=GNZIAbf0gT0

https://leetcode.com/problems/can-i-win/discuss/95277/Java-solution-using-HashMap-with-detailed-explanation




https://www.youtube.com/watch?v=md3qQ-5B0aU


class Solution {
    public boolean canIWin(int max, int total) {
        // pruning 
        if((1 + max) * max / 2  < total) return false;
        
        int[] state = new int[max + 1];
        Map<String, Boolean> map = new HashMap<>();
        return dfs(state, total, map);
    }
    private boolean dfs(int[] state, int total, Map<String, Boolean> map){
        String key = Arrays.toString(state); // can't use new String() here 
        if(map.containsKey(key)) return map.get(key);
        
        for(int i = 1; i < state.length; i++){
            if(state[i] == 0){
                state[i] = 1;
                // if the first player can win , then i > the total left
                // if the second player cann't win, 
                if(total - i <= 0 || !dfs(state, total - i, map)){
                    map.put(key, true);
                    // backtrack 
                    state[i] = 0;
                    return true;
                }
                // back track 
                state[i] = 0;
            }
        }
        // if we haven't returned true yet 
        map.put(key, false);
        return false;
    }
}