leetcode486. Predict the Winner

王军发布于2019-08-19 10:25 / 435人阅读

摘要：但是，往往会有可以优化的空间。假设我们用来记录子数组之间，第一个取数字的玩家和第二个取数字的玩家之间最大的差距。再考虑初始情况，即当数组长度为时，可以得知此时玩家一和玩家二之间的差距即为该数组元素的值。

题目要求

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

假设有一个正整数数组，两名玩家轮流从里面取数组，玩家1先取，玩家2后取，要求判断出玩家1是否一定能够取胜？

思路和代码

看到这种题目的时候，会直观的想到，如果我能够暴力的遍历出玩家1和玩家2之间所有的取数字的方式，就一定可以算出玩家1是否能够取胜。但是，往往会有可以优化的空间。假设我们用diffi来记录子数组i~j之间，第一个取数字的玩家和第二个取数字的玩家之间最大的差距。则 diffi = Math.max(nums[i]-diffi+1, nums[j+1]-diffi), 即从左取第一个数字或是从右取第一个数字能够获得的最大差距。再考虑初始情况，即当数组长度为1时，可以得知此时玩家一和玩家二之间的差距即为该数组元素的值。代码如下：

    public boolean PredictTheWinner(int[] nums) {
        int[][] diff = new int[nums.length][nums.length];
        for(int i = 0 ; i= 0;
    }

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