https://leetcode.com/problems/can-i-win/#/description 是两个人互相取数字, 当第一个人取的数字超过目标, 就return true 原本的想法是, player 1 挑全部没选过的number, 然後 呼叫secondPlayerWin的 function 去判断是不是有存在secondPlayer win的, 只要有存在A 选的这个number就是不行的 不过写不太好的吃了个wrong answer, 偷看看讨论串解答 看了很多的作法, 都是做类似 !helper(desiredTotal - i) 的递回, 想半天仍然不太懂... 有版友有兴趣一起研究研究吗? 这个是原作者的解释, 但是我仍然不懂他的意思, 为什麽code要写成那样 ** The strategy is we try to simulate every possible state. E.g. we let this player choose any unchosen number at next step and see whether this leads to a win. If it does, then this player can guarantee a win by choosing this number. If we find that whatever number s/he chooses, s/he won't win the game, then we know that s/he is guarantee to lose given such a state. // try every unchosen number as next step for(int i=1; i<used.length; i++){ if(!used[i]){ used[i] = true; // check whether this lead to a win, which means helper(desiredTotal-i) must return false (the other player lose) if(!helper(desiredTotal-i)){ map.put(key, true); used[i] = false; return true; } used[i] = false; } } map.put(key, false); -- If people do not believe that mathematics is simple, it is because they do not realize how complicated life is. -- John Louis von Neumann --

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