[leetcode] 486. Predict the Winner

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最新推荐文章于 2025-05-22 02:00:00 发布

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分类专栏： leetcode C++ leetcode题解

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本文介绍了一个关于两个玩家轮流从数组两端选择非负整数的博弈问题，并使用动态规划来预测第一个玩家是否能赢得比赛。通过递归算法实现，详细解释了如何计算每个玩家的最优得分策略。

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Description

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.

分析

题目的意思是：两个玩家从数组里面轮流拿数字，只能从两头取，每个人都拿数都是最优的，问最后第一个玩家能赢吗？

这里需要用到动态规划，动态规划其实可以转换为递归，下面就是递归。
dp[i][j] = max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1]); dp[i][j]第一个玩家拿了第i个数，第二个玩家拿第j个数后，第一个玩家比第二个玩家多的分值，使它最大化。

代码

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        int low=0;
        int high=nums.size()-1;
        return solve(nums,0,nums.size()-1)>=0;
    }
    
    int solve(vector<int>& nums,int s,int e){
        if(s==e){
            return nums[e];
        }
        return max(nums[s]-solve(nums,s+1,e),nums[e]-solve(nums,s,e-1));
    } 
};