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.

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：这题跟 [leetcode] Stone Game 是一模一样的题目；代码完全一样；

区间型动态规划，f[i][j] 表示的物理意义是：面对i 到j的石头，下棋子的人，我可以拿到的最大的与对手的数字差

f[i][j] = max {a[i] - f[i + 1][j], a[j] - f[i][j - 1]} , 最后如果f[0][n - 1] >= 0 代表先手alex能够拿到正数，也就是比lee拿得多，从而赢；

初始化： f[i][i] = a[i],  就面对一个数字；

计算顺序：

长度1： f[0][0] f[1][1] f[2][2]....f[n - 1][n - 1]

长度2：f[0][1],.....f[n - 2][n-1]

...

长度N：f[0][n - 1]

class Solution {
    public boolean PredictTheWinner(int[] A) {
        if(A == null || A.length == 0) {
            return false;
        }
        int n = A.length;
        int[][] f = new int[n][n];
        
        for(int i = 0; i < n; i++) {
            f[i][i] = A[i];
        }
        
        for(int len = 2; len <= n; len++) {
            for(int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                f[i][j] = Math.max(A[i] - f[i + 1][j], A[j] - f[i][j - 1]);
            }
        }
        return f[0][n - 1] >= 0;
    }
}