486. Predict the Winner

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.
就是2个player轮流取一个数组两端的数，博弈，看谁最终SUM最大就赢了。给定一个字符串，判断player1能不能赢。
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.

思路

存在博弈，最优解/次优解所以是动态规划。
本题难点是找到递推方程，dp[i][j]是每次第一个人可以赢的差值。
所以
Your input

[1,5,2]
[1,5,233,7]
[1,5,233,7,9]

Your stdout

1, 4, -2, 
0, 5, 3, 
0, 0, 2, 

1, 4, 229, 222, 
0, 5, 228, -221, 
0, 0, 233, 226, 
0, 0, 0, 7, 

1, 4, 229, 222, -213, 
0, 5, 228, -221, 230, 
0, 0, 233, 226, 231, 
0, 0, 0, 7, 2, 
0, 0, 0, 0, 9, 

可以看出，

每个dp[i][j] = max(字符串左端nums[i] - dp表下面值, 字符串右端nums[j] - dp表左面值)

代码

class Solution {
    public boolean PredictTheWinner(int[] nums) {
        if(nums==null || nums.length==0) return true;
        int n = nums.length;
        int[][] dp=new int[n][n];
        for(int i=n-1;i>=0;i--){
            dp[i][i]=nums[i];
            for(int j=i+1;j<n;j++){
                dp[i][j] = Math.max(nums[i]-dp[i+1][j],nums[j]-dp[i][j-1]);
            }
        }
        /*
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                System.out.print(dp[i][j]+ ", ");
            }
            System.out.println();
        }
        */
        return dp[0][n-1]>=0;
    }
}