Retrograde Method: http://codepad.org/i3CsFKnT (Perl)
不知道有没有写错
还请各位帮我看看
※ 引述《powertodream (The Beginning)》之铭言：
: 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, 然后 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, 为什么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);