486. Predict the Winner

最新推荐文章于 2019-05-16 00:03:41 发布

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最新推荐文章于 2019-05-16 00:03:41 发布

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分类专栏： LeetCode

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本文介绍了一种预测游戏中哪位玩家能赢得比赛的算法。通过递归的方式模拟两位玩家轮流从数组两端取数的过程，并利用回溯算法来评估每种选择的结果。最终确定先手玩家是否有获胜的可能性。

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

结合Can I Win思路求解，程序如下所示：

class Solution {
    public boolean PredictTheWinner(int[] nums) {
        if (nums.length == 1){
            return true;
        }
        return canIWin(nums, 0, nums.length - 1, 0, 0);
    }
    public boolean canIWin(int[] nums, int left, int right, int player1, int player2){
        if (left > right){
            return player1 >= player2;
        }
        player1 += nums[left];
        left ++;
        if (!canIWin(nums, left, right, player2, player1)){
            return true;
        }
        left --;
        player1 -= nums[left];
        
        player1 += nums[right];
        right --;
        if (!canIWin(nums, left, right, player2, player1)){
            return true;
        }
        right ++;
        player1 -= nums[right];
        return false;
    }
}