Leetcode 486. Predict the Winner 预测赢家解题报告

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博客介绍了LeetCode 486问题的解题思路，玩家在数组中依次选择头部或尾部的数字作为分数，目标是判断Player1是否有必胜策略。通过递归方法确定是否存在一种选择能让Player1确保胜利，关键在于每次选择的分数要大于对手可能获得的分数。

1 解题思想

开学了，我又回归了。
这道题的意思是给了一个数组，两个人玩，每个人可以在当前的数组的头部或者尾部选择一个数，作为自己的分数，然后换人，已经被选过的数字不能再用。
现在有两个人玩，Player1和Player2，问给了当前的数组后，Player1能不能保证胜利？

解题方法，就是需要做一个决策，看Player1有没有一个必胜的选择决策，这个只需要递归下去选就好。
所谓的必胜，就是当前选到的分数，大于对手能选择到的。

2 原题

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.

3 AC解

public class Solution {
    public boolean PredictTheWinner(int[] nums) {
        return helper(nums, 0, nums.length-1)>=0;
    }

    private int helper(int[] nums, int start, int end){  
        if(start==end){
            return nums[start];
        } else{
            return Math.max(nums[start] - helper(nums,start+1,end),nums[end] - helper(nums,start,end-1));
        }
    }
}