动态规划——486. Predict the Winner［Medium］

题目描述

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.

两个人从数组头或尾依次抽取数字，两人都希望自己取得最大和，问首先取的那个人是否能赢。

解题思路

这和课本上的不一样，因为从example2看出，player1并没有使用贪心算法！（因为为了拿到233，他一开始先拿了1而不是更大的7）

博客里有dalao提到用MiniMax，但是很遗憾，解题思路让我有些不太理解，而且状态转移并不是很容易想到，于是我采取了另一种方法，感谢Shilei Tian的思路！

——模拟两人的游戏过程：

1、bool first(vector<int>& num, int i, int j, int p1, int p2)表示第一个玩家的结果

2、bool second(vector<int>& num, int i, int j, int p1, int p2)表示第二个玩家的结果

3、i、j表示可选择的数字首尾，i > j 此时已经没有能选择的了，返回结果，不然递归调用，而且first的结果与second是相反关系，first ＝ ！second

代码如下

bool second(vector<int>& num, int i, int j, int p1, int p2);bool first(vector<int>& num, int i, int j, int p1, int p2) {if (i > j)return p1 >= p2;return !second(num, i + 1, j, p1 + num[i], p2) || !second(num, i, j - 1, p1 + num[j], p2);}bool second(vector<int>& num, int i, int j, int p1, int p2) {if (i > j)return p2 > p1;return !first(num, i + 1, j, p1, p2 + num[i]) || !first(num, i, j - 1, p1, p2 + num[j]);}class Solution {public:    bool PredictTheWinner(vector<int>& nums) {        return first(nums, 0, nums.size() - 1, 0, 0);    }};

很遗憾，这并不像动态规划解法，但是却很容易理解。