Predict the Winner

题目：

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.

题目大意：

给定一个非负整数数组表示一组分数。玩家1从数组两端任取一个数字，之后由玩家2选取，以此类推。每当一名玩家选择一个数字之后，另一位玩家就不能再选这个数字。重复此过程直到所有的数字都被选完。分数高的玩家获胜。

给定分数数组，预测玩家1是否可以获胜。可以假设每一个玩家都采用最优策略游戏。

注意：

1. 数组长度∈[1, 20]
2. 任意分数均为非负整数，且不超过10,000,000
3. 如果分数相同，则判定为玩家1获胜

解题思路：

   这道题应该用递归的方法来做，当前玩家赢返回true的条件就是递归调用下一个玩家输返回false。我们需要一个变量来标记当前是第几个玩家，还需要两个变量来分别记录两个玩家的当前数字和，在递归函数里面，如果当前数组为空了，我们直接比较两个玩家的当前得分即可，如果数组中只有一个数字了，根据玩家标识来将这个数字加给某个玩家并进行比较总得分。如果数组有多个数字，分别生成两个新数组，一个是去掉首元素，一个是去掉尾元素，然后根据玩家标识分别调用不同的递归，只要下一个玩家两种情况中任意一种返回false了，那么当前玩家就可以赢了。

代码如下：

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        return canWin(nums, 0, 0, 1);
    }
    bool canWin(vector<int> nums, int sum1, int sum2, int player) {
        if (nums.empty()) return sum1 >= sum2;
        if (nums.size() == 1) {
            if (player == 1) return sum1 + nums[0] >= sum2;
            else if (player == 2) return sum1 + nums[0] < sum2;
        }
        vector<int> va = vector<int>(nums.begin() + 1, nums.end());
        vector<int> vb = vector<int>(nums.begin(), nums.end() - 1);
        if (player == 1) {
            return !canWin(va, sum1 + nums[0], sum2, 2) || !canWin(vb, sum1 + nums.back(), sum2, 2);
        } else if (player == 2) {
            return !canWin(va, sum1, sum2 + nums[0], 1) || !canWin(vb, sum1, sum2 + nums.back(), 1);
        }
    }
};