486. Predict the Winner

486. Predict the Winner

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• Total Accepted: 10743 
• Total Submissions: 24072 
• Difficulty: Medium
• Contributors: sameer13

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: FalseExplanation: 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: TrueExplanation: 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.

解题思路：

用dp[i][j]记录从第i到第j个数，先行玩家比另一玩家多得的分数。最终只要dp[0][nums.size()-1]＞0，玩家1就能够取胜。

每个玩家都能够从数组的左端或者右端取数，假设目前剩余的是从i到j位的数组，轮到玩家A取数。如果A取了左边的数，即i，那么A能够立刻获得nums[i]分。下一步轮到B从i+1到j中选择。注意这时候，dp[i+1][j]这个时候已经记录了从第i+1到第j个数，先行玩家比另一玩家多得的分数，所以在i+1到j，B最终能够比A多得dp[i+1][j]分。因此，如果A选择了i，那么当他们选择完所有的数之后A最多能比B多得nums[i]-dp[i+1][j]分。相似的，如果A取了右边的数，即j，那么A最多能比B多得nums[j]-dp[i][j-1]分。因此可以得出状态转移方程：

dp[i][j] = max(nums[i] - dp[i + 1][j], nums[j] - dp[i][j - 1])

容易得知dp[i][i] = nums[i]

代码：

#include<math>class Solution {public:    bool PredictTheWinner(vector<int>& nums) {        int dp[20][20];        for(int i = 0; i < nums.size(); i++)        dp[i][i] = nums[i];        for(int len = 1; len < nums.size(); len++){        for(int i = 0; i < nums.size() - len; i++){        int j = i + len;        dp[i][j] = max(nums[i] - dp[i+1][j], nums[j] - dp[i][j-1]);        }        }        return dp[0][nums.size()-1] >= 0;    }};