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.
 

Constraints:

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.

一、递归搜索，博弈树模型

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        return total(nums,0,nums.size()-1,1) >=0;
    }
    int total(vector<int>& nums,int start,int end,int turn)
    {
        if(start == end){
            return nums[start]*turn;
        }
        int scoreStart = nums[start]*turn +total(nums,start+1,end,-turn);
        int scoreEnd = nums[end]*turn + total(nums,start,end-1,-turn);
        return max(scoreEnd*turn , scoreStart*turn) *turn;//极大极小值搜索，先手取最大的那个数，后手去最小的那个数
    }
};

注意：

1.数组是连续的，可以用start,end标记递归下去。

2.极大极小值搜索，为了确定是否能赢，在第一方选择是注意选择最大的，在第二方选择时，考虑到最坏的情况，选择最小的，保证在最坏的情况下，第一方也能胜利。return max(scoreEnd*turn,scoreStart*turn)*turn。

3.也可以进行剪枝，但是难度增加

二、动态规划

 

定义二维数组 dp[i][j],其行数和列数都等于数组的长度，dp[i][j],表示从i到j到最大分差，注意当前玩家不一定是先手。

i =j  dp[i][i] =nums[i];

i<j dp[i][j] =max(nums[i] -nums[i+1][j],nums[j]-nums[i][j-1]);

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {//动态规划
        int n =nums.size();
        vector<vector<int>> dp(n,vector<int>(n,0));
        for(int i=0;i<n;++i){
            dp[i][i] = nums[i];
        }
        for(int i = n-2;i>=0;i--){
            for(int j=i+1;j<n;++j){
                dp[i][j] = max(nums[i] - dp[i+1][j],nums[j] -dp[i][j-1]);
            }
        }
        return dp[0][n-1] >= 0;
    }
};

三、动态规划压缩空间

由于只用到i+1和第i行，且第i行的为j-1，即左侧更新后的又被用到，所以遍历顺序不变，直接压缩空间

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