464. Can I Win

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.

动态规划解题。动规的规则是要保证第一位赢，就要保证第二位输。那么先得到下面的代码：

public class Solution {    public boolean canIWin(int maxChoosableInteger, int desiredTotal) {        if (((maxChoosableInteger + 1) * maxChoosableInteger) / 2 < desiredTotal) {            return false;        }        if (desiredTotal <= 0) {            return true;        }        boolean[] used = new boolean[maxChoosableInteger + 1];        return helper(used, desiredTotal);    }        private boolean helper (boolean[] used, int desiredTotal) {        if (desiredTotal <= 0) {            return false;        }            for (int i = 1; i < used.length; i ++) {                if (!used[i]) {                    used[i] = true;                    if (!helper(used, desiredTotal - i)) {                        used[i] = false;                        return true;                    }                    used[i] = false;                }            }        }        return false;    }}

但是这种方法会超时，需要优化。优化思路是用hashmap保存之前的策略，如果hashmap有的话，直接返回结果，没有的话再更新。代码如下：

public class Solution {    Map<Integer, Boolean> map;        public boolean canIWin(int maxChoosableInteger, int desiredTotal) {        if (((maxChoosableInteger + 1) * maxChoosableInteger) / 2 < desiredTotal) {            return false;        }        if (desiredTotal <= 0) {            return true;        }        boolean[] used = new boolean[maxChoosableInteger + 1];        map = new HashMap<Integer, Boolean>();        return helper(used, desiredTotal);    }        private boolean helper (boolean[] used, int desiredTotal) {        if (desiredTotal <= 0) {            return false;        }        int key = transfer(used);        if (!map.containsKey(key)) {            for (int i = 1; i < used.length; i ++) {                if (!used[i]) {                    used[i] = true;                    if (!helper(used, desiredTotal - i)) {                        map.put(key, true);                        used[i] = false;                        return true;                    }                    used[i] = false;                }            }            map.put(key, false);        }        return map.get(key);    }        private int transfer(boolean[] used) {        int ret = 0;        for (boolean use: used) {            ret = ret << 1;            if (use) {                ret |= 1;            }        }        return ret;    }}

hashmap还不够快，就用一个int来存储状态。代码如下：

public class Solution {     public boolean canIWin(int maxChoosableInteger, int desiredTotal) {        if(maxChoosableInteger<1) return false;        if(desiredTotal<1) return true;        if(maxChoosableInteger*(1+maxChoosableInteger)/2<desiredTotal) return false;        /* We will use int to record the state of usage of integers from pool [1,maxChoosableInteger]        For example: maxChoosableInteger=2, we can choose from [1,2]        we use 0b00 to represent no number has been used               0b01 to represent 1 has been used               0b10 to represent 2 has been used               0b11 to represent 1,2 has been used        */        /* memo[i] used to record under the situation of state i, whether the player can force a win           memo[i]=0 means we have no information for this state                   1 means we can force a win under state i                  -1 means we can't force a win under state i        */        int[] memo=new int[1<<maxChoosableInteger];        return dfs(0, 0, memo, maxChoosableInteger,desiredTotal);    }    public boolean dfs(int state,int currentSum, int[] memo, int max, int desiredTotal){        if(memo[state]!=0) return memo[state]==1;        if(currentSum>=desiredTotal){            memo[state]=-1;            return false;        }        /*By pick one element from available element, search for if there is a choice such that we can force a win        */        for(int i=0;i<max;i++){            if(((state>>i)&1)==0){//number i+1 is available                if(!dfs(state|(1<<i), currentSum+i+1, memo, max, desiredTotal)){                    memo[state]=1;                    return true;                }            }        }        memo[state]=-1;        return false;    }}