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  • 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 = 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.

    or this question, the key part is: what is the state of the game? Intuitively, to uniquely determine the result of any state, we need to know:

    1. The unchosen numbers
    2. The remaining desiredTotal to reach

    A second thought reveals that 1) and 2) are actually related because we can always get the 2) by deducting the sum of chosen numbers from original desiredTotal.

    Then the problem becomes how to describe the state using 1).

     For this problem, by applying the memo, we at most compute for every subproblem once, and there are O(2^n) subproblems, so the complexity is O(2^n) after memorization. (Without memo, time complexity should be like O(n!))

    public class WinGame {
    /*
     * 
     * 
     * boolean canIWin(int maxNum, int target),
     * 从1,2...maxNum的数组里两个玩家轮流选数,
     * 第一个达到sum>=target的玩家获胜,问如何判断先选的玩家能获胜。
     * sum是指连个玩家一起选的数的和
     * 
     */
    	
    	public static boolean canWin(int maxNum, int target) {
    		boolean[] isUsed = new boolean[maxNum + 1];
    		return helper(isUsed, target, maxNum);
    	}
    	
    	public static boolean helper(boolean[] isUsed, int target, int maxNum) {
    		if (maxNum < 0) {
    			return false;
    		}
    		boolean isEmpty = true;
    		for (boolean used : isUsed) {
    			if (!used) {
    				isEmpty = false;
    			}
    		}
    		if (isEmpty) {
    			return false;
    		}
    		if (target <= 0) {
    			return false;
    		}
    		boolean canWin = false;
    		for (int i = 1; i <= maxNum; i++) {
    			if (!isUsed[i] && i >= target) {
    				return true;
    			}
    			if (!isUsed[i]) {
    				isUsed[i] = true;
    				canWin = canWin || !helper(isUsed, target - i, maxNum);
    				isUsed[i] = false;
    			}
    		}
    		return canWin;
    	}
    	
    	public static boolean canIWin(int[] numberPool, int target) {
            boolean isEmpty = true;
            for (int data : numberPool)
                if (data > 0) isEmpty = false; 
            if (isEmpty) return false;
            else {
                if (target <= 0) return false;
                for (int data : numberPool)
                    if (data > 0 && data >= target) return true;
                boolean canIWinFlag = false;
                for (int i = 0; i < numberPool.length && numberPool[i] > 0; ++i) {
                    int data = numberPool[i];
                    numberPool[i] = -1;
                    canIWinFlag = canIWinFlag || !canIWin(numberPool, target - data); // other's turn
                    numberPool[i] = data;
                }
                return canIWinFlag;
            }
        }
    	
    	public static void main(String[] args) {
    		int[] a = {1, 2, 3, 4, 5, 6, 7};
    		//System.out.print(canIWin(a, 17));
    		System.out.print(canWin(6, 17));
    	}
    }
    

      

    public class Solution {
        Map<String, Boolean> map;
        boolean[] used;
        public boolean canIWin(int maxChoosableInteger, int desiredTotal) {
            int sum = (1+maxChoosableInteger)*maxChoosableInteger/2;
            if(sum < desiredTotal) return false;
            if(desiredTotal <= 0) return true;
            
            map = new HashMap();
            used = new boolean[maxChoosableInteger+1];
            return helper(desiredTotal);
        }
        
        public boolean helper(int desiredTotal){
            if(desiredTotal <= 0) return false;
            String key = Arrays.toString(boolean[]);
            if(!map.containsKey(key)){
        // 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 (i.e. the other player lose)
                        if(!helper(desiredTotal-i)){
                            map.put(key, true);
                            used[i] = false;
                            return true;
                        }
                        used[i] = false;
                    }
                }
                map.put(key, false);
            }
            return map.get(key);
        }
    }
    

      

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  • 原文地址:https://www.cnblogs.com/apanda009/p/7946311.html
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