486. Predict the Winner

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: 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.
Example 2:
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.
Note:
1 <= length of the array <= 20.
Any scores in the given array are non-negative integers and will not exceed 10,000,000.
If the scores of both players are equal, then player 1 is still the winner.
这道题就是要分4种情况进行讨论，分4个vector容器去看。
代码如下：

class Solution {public:    bool PredictTheWinner(vector<int>& nums) {        int sums=0;        for(int i=0;i<nums.size();i++){            sums+=nums[i];        }        int maxx=MMax(nums);        return maxx>=sums-maxx;    }    int MMax(vector<int>& nums){        if(nums.size()==1)return nums[0];        if(nums.size()==2)return max(nums[0],nums[1]);        else{            vector<int>v1(nums);            vector<int>v2(nums);            vector<int>v3(nums);            int begin=nums[0];            int end=nums[nums.size()-1];            v1.erase(v1.begin());            v1.erase(v1.begin());            v2.erase(v2.begin());            v2.erase(v2.end()-1);            v3.erase(v3.end()-1);            v3.erase(v3.end()-1);            return max(begin+min(MMax(v1),MMax(v2)),end+min(MMax(v2),MMax(v3)));        }    }};

注意erase函数的应用还有vectorv1(nums)这种方式的直接copy函数调用。还有erase(v3.end()-1)而不是erase(v3.end())。