// 486. Predict the Winner #include "leetcode.h" /* * given an integer array 'nums'. two players are playing a game with this * array: player 1 and player 2. player 1 and player 2 take turns, with player 1 * starting first. both players start the game with a score of zero. at each * turn, the player takes one of the numbers from either end of the array (ie. * 'nums[0]', or 'nums[nums.len - 1]') which reduces the size of the array by * one. the player adds the chosen number to their score. the game ends when * there are no more elements in the array. return 'true' if player 1 can win * the game. if the scores of both players are equal, then player 1 is still the * winner, and you should also return 'true'. you may assume that both players * are playing optimally. */ class Solution { bool check_win(int ans, int total) { return ans >= total - ans; } int max_score(vector &nums, int total, int i, int j) { if (i > j) return 0; return total - min(max_score(nums, total - nums[i], i + 1, j), max_score(nums, total - nums[j], i, j - 1)); } public: bool PredictTheWinner(vector &nums) { int total = 0; for (auto i : nums) total += i; return check_win(max_score(nums, total, 0, nums.size() - 1), total); } }; int main() { Solution obj; vector n1 = {1, 5, 2}, n2 = {1, 5, 233, 7}; printf("%d\n", obj.PredictTheWinner(n1)); // expect: 0 printf("%d\n", obj.PredictTheWinner(n2)); // expect: 1 }