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.

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.

题解：
两个player, 从头部或者尾部取数字，然后follow by第二个。最后分数最高的获胜。求问最后谁会获胜。

用DP[i][j]表示，player1可以从Array[i][j]获得比player2获得多的分数。空间复杂度O(n^{2)，时间复杂度O(n}2)。

public boolean PredictTheWinner(int[] nums) {
    int n = nums.length;
    int[][] dp = new int[n][n];
    for (int i = 0; i < n; i++) { dp[i][i] = nums[i]; }
    for (int len = 1; len < n; len++) {
        for (int i = 0; i < n - len; i++) {
            int j = i + len;
            dp[i][j] = Math.max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1]);
        }
    }
    return dp[0][n - 1] >= 0;
}

优化：O(N) space complexity:

public boolean PredictTheWinner(int[] nums) {
    if (nums == null) { return true; }
    int n = nums.length;
    if ((n & 1) == 0) { return true; } // Improved with hot13399's comment.
    int[] dp = new int[n];
    for (int i = n - 1; i >= 0; i--) {
        for (int j = i; j < n; j++) {
            if (i == j) {
                dp[i] = nums[i];
            } else {
                dp[j] = Math.max(nums[i] - dp[j], nums[j] - dp[j - 1]);
            }
        }
    }
    return dp[n - 1] >= 0;
}