### LeetCode  • ㊗️
• 大家
• offer
• 多多！

## Problem

In the “100 game” two players take turns adding, to a running total, any integer from 1 to 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 from 1 to 15 without replacement until they reach a total >= 100.

Given two integers maxChoosableInteger and desiredTotal, return true if the first player to move can force a win, otherwise, return false. Assume both players play optimally.

Example 1:

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.


Example 2:

Input: maxChoosableInteger = 10, desiredTotal = 0
Output: true


Example 3:

Input: maxChoosableInteger = 10, desiredTotal = 1
Output: true


Constraints:

• 1 <= maxChoosableInteger <= 20
• 0 <= desiredTotal <= 300

## Code

class Solution {
int[] memo;
public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
int sum = (1 + maxChoosableInteger) * maxChoosableInteger / 2;
if(sum < desiredTotal) return false;
if(desiredTotal <= 0) return true;

return helper(maxChoosableInteger, desiredTotal, 0, new HashMap<>());
}

private boolean helper(int maxNum, int target, int state, HashMap<Integer, Boolean> map){
if(target <= 0) return false;
if(map.containsKey(state)) return map.get(state);

for(int i = 0; i < maxNum; i++){
if((state & (1 << i)) != 0) continue;

if(!helper(maxNum, target - (i + 1), state | (1 << i), map)){
map.put(state, true);
return true;
}
}

map.put(state, false);
return false;
}
}