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.
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.
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.
玩家P1，P2轮流从数组两端选择一个数字，最后拿到数字的和多的那一方胜。P1先拿，判断玩家1是否能胜利。
解决：创建一个n*n的数组dp，因为是P1先拿，所以dp[0][n-1]或dp[n-1][0]就是玩家1能够获得的分数。
假设有数组nums[i: j]，当玩家取nums[i]，那么dp[i][j] = nums[i] + sum[i+1][j] - max(dp[i+1][j], dp[j][i+1]);
当玩家取nums[j]，那么dp[j][i] = nums[j] + sum[i][j-1] - max(dp[i][j+1], dp[j+1][i]);
最后要注意一下数组的更新顺序。
class Solution { public: bool PredictTheWinner(vector<int>& nums) { int len = nums.size(); vector<vector<int>> sum(len, vector<int>(len)); for (int i=0; i<len; ++i) sum[i][i] = nums[i]; for (int i=0; i<len; ++i) for (int j=i+1; j<len; ++j) { sum[i][j] = sum[i][j-1] + nums[j]; sum[j][i] = sum[i][j]; } vector<vector<int>> dp(len, vector<int>(len)); for (int i=0; i<len; ++i) dp[i][i] = nums[i]; for (int i=1; i<len; ++i) { for (int j=0; j<i; ++j) dp[j][i] = nums[i] + sum[j][i-1] - max(dp[j][i-1], dp[i-1][j]); for (int j=i-1; j>=0; --j) dp[i][j] = nums[j] + sum[j+1][i] - max(dp[j+1][i], dp[i][j+1]); } return 2*max(dp[len-1][0], dp[0][len-1]) >= sum[len-1][0]; } };