486. Predict the Winner_registry winner注册码-优快云博客

本文链接：https://blog.youkuaiyun.com/BigFatSheep/article/details/84243459

本文介绍了一个策略游戏的解决方案，通过动态规划预测两个玩家谁将赢得比赛。游戏规则为两名玩家从数组两端轮流取数，目标是获得比对手更高的总分。文章详细解释了如何使用二维数组进行动态规划，并提供了具体示例和代码实现。

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Description

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.

Problem URL

Solution

给row 和 col，给一个String数组，问这个数组能在这个显示器上show 出几遍。

Using a two dimensional array to perform dynamic programming. Fill dp with -1 first. Because Arrays.fill can’t assign value to multi dimensional array, we should do it row by row. Then calculate the sum of the whole nums for future judgement.

Because when the number is a even number, first taker will always win. Because he could always leave a worse choice for the second player.

For the predict helper, the dp array stores how much score first player could make after one round. One round is that first player takes one number and second player takes one number too. Their position are I and j. If we have dp value , return it. Or calculate different circumstance and take the bigger one.

Code

class Solution {
    public boolean PredictTheWinner(int[] nums) {
        if (nums.length % 2 == 0){
            return true;
        }
        
        int n = nums.length;
        int[][] dp = new int[n][n];
        for (int[] row : dp){
            Arrays.fill(row, -1);
        }
        
        int sum = 0;
        for (int i : nums){
            sum += i;
        }
        int score = predictHelper(nums, dp, 0, n - 1);
        return 2 * score >= sum;
    }
    
    private int predictHelper(int[] nums, int[][] dp, int i, int j){
        if (i > j){
            return 0;
        }
        if (dp[i][j] != -1){
            return dp[i][j];
        }
        
        int a = nums[i] + Math.min(predictHelper(nums, dp, i + 1, j - 1),predictHelper(nums, dp, i + 2, j));
        int b = nums[j] + Math.min(predictHelper(nums, dp, i, j - 2),predictHelper(nums, dp, i + 1, j - 1));
        dp[i][j] = Math.max(a, b);
        
        return dp[i][j];
    }
}