464 Can I Win

最新推荐文章于 2019-01-22 05:10:11 发布

原创最新推荐文章于 2019-01-22 05:10:11 发布 · 300 阅读

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本文探讨了两人猜数游戏的一种变种玩法，在这种变种中，玩家从指定范围内选择不可重复使用的数字，以使总和达到或超过目标值。文章提供了详细的算法实现，包括如何判断先手玩家是否能够确保胜利。

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464 Can I Win

题目描述：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.

题目大意：两个人猜数，给定两个数，一个代表可以选择的范围maxChoosableInteger，一个代表目标数字desiredTotal，每个人可以从maxChoosableInteger选择数字，但是数字不可重复选择，给定maxChoosableInteger与desiredTotal判断第一个人是否能获胜

代码

 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;
//                另一个人不能赢
              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;
  }