Leetcode 464. Can I Win
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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.
首先做两个基本判断, (1)如果 maxChooseInteger 大于或等于 desiredTotal,返回正确。
(2)如果 所有数的和加起来 小于 desiredTotal, 返回错误。
然后,使用归递来解决。用一个 HashMap 来储存 状态 和 结果,
public boolean canIWin(int maxChoosableInteger, int desiredTotal) { if (maxChoosableInteger >= desiredTotal) return true; if (maxChoosableInteger * (1 + maxChoosableInteger) / 2 < desiredTotal) return false; return helper(desiredTotal, new int[maxChoosableInteger], new HashMap<>()); } private boolean helper(int total, int[] state, HashMap<String, Boolean> map) { String cur = Arrays.toString(state); if (map.containsKey(cur)) return map.get(cur); for (int i = 0; i < state.length; i++) { if (state[i] == 0) { 如果没有访问 state[i] = 1; 记录下访问过 if (total <= i + 1 || !helper(total - (i + 1), state, map)) { total 小于 maxChoosableInteger map.put(cur, true); state[i] = 0; return true; } state[i] = 0; } } map.put(cur, false); return false; }
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