486. Predict the Winner

最新推荐文章于 2019-05-16 00:03:41 发布

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最新推荐文章于 2019-05-16 00:03:41 发布

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分类专栏： LeetCode Dynamic Programming MiniMax 文章标签： LeetCode Dynamic Programming miniMax Predict the Winner

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/71799237

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本文介绍了一种通过算法预测两位玩家在游戏中胜负的方法。游戏规则为双方轮流从整型数组两端取值，最终比较总和确定胜者。文章详细解释了递归与动态规划的解决方案，并提供了具体的代码实现。

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Given an array of scores that are non-negative integers. Player 1 picks one of the numbers from either end of the array followed by the player 2 and then player 1 and so on. Each time a player picks a number, that number will not be available for the next player. This continues until all the scores have been chosen. The player with the maximum score wins.

Given an array of scores, predict whether player 1 is the winner. You can assume each player plays to maximize his score.

Example 1:

Input: [1, 5, 2]
Output: False
Explanation: Initially, player 1 can choose between 1 and 2. 
If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. If player 2 chooses 5, then player 1 will be left with 1 (or 2). 
So, final score of player 1 is 1 + 2 = 3, and player 2 is 5. 
Hence, player 1 will never be the winner and you need to return False.

Example 2:

Input: [1, 5, 233, 7]
Output: True
Explanation: Player 1 first chooses 1. Then player 2 have to choose between 5 and 7. No matter which number player 2 choose, player 1 can choose 233.
Finally, player 1 has more score (234) than player 2 (12), so you need to return True representing player1 can win.

Note:

1 <= length of the array <= 20.
Any scores in the given array are non-negative integers and will not exceed 10,000,000.
If the scores of both players are equal, then player 1 is still the winner.

对于这道题，双方都在pick数，最后比大小，可以转化成p1的操作是+，p2的操作是-。动规的原则是return s==e ? nums[e] : Math.max(nums[e] - helper(nums, s, e-1), nums[s] - helper(nums, s+1, e))。代码如下：

public class Solution {
    public boolean PredictTheWinner(int[] nums) {
        return helper(nums, 0, nums.length-1)>=0;
    }
    private int helper(int[] nums, int s, int e){        
        return s==e ? nums[e] : Math.max(nums[e] - helper(nums, s, e-1), nums[s] - helper(nums, s+1, e));
    }
}

可以加一个cache优化代码效率，代码如下：

public class Solution {
    public boolean PredictTheWinner(int[] nums) {
        return helper(nums, 0, nums.length - 1, new int[nums.length][nums.length]) >= 0;
    }
    
    private int helper(int[] nums, int s, int e, int[][] memo) {
        if (memo[s][e] == 0) {
            memo[s][e] = s == e? nums[e]: Math.max(nums[e] - helper(nums, s, e - 1, memo), nums[s] - helper(nums, s + 1, e, memo));
        }
        return memo[s][e];
    }
}