486. Predict the Winner Medium

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.

思路：使用动态规划思维，子问题为：在一个更小的子列中player1能否获胜。用二维数组sum[i][j]记录从下标i到j先拿的人可以比后拿的人多的分数，只要最终sum[0][nums.size() -1] >= 0,那么player1就赢。

class Solution {public:    bool PredictTheWinner(vector<int>& nums) {        int sum[nums.size()][nums.size()];        for (int i = 0; i < nums.size(); i++) {            sum[i][i] = nums[i];        }        for (int i = nums.size() - 2; i >= 0; i--) {            for (int j = i + 1; j < nums.size(); j++) {                sum[i][j] = max(nums[i] - sum[i + 1][j], nums[j] - sum[i][j - 1]);            }        }        if (sum[0][nums.size() - 1] >= 0) {            return true;        }        return false;    }};