[leetcode]486. Predict the Winner

题目链接：https://leetcode.com/problems/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. 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.

方法一（超时）：

class Solution{
public:
    bool PredictTheWinner(vector& nums)
    {
        //vector[start][end]代表nums的头索引为start,尾索引为end时player1得到的最大的score
        vector> scores(20,vector(20,0));
        int sum=accumulate(nums.begin(),nums.end(),0);
        int target=(sum%2)?sum/2+1:sum/2;
        return maxScore(nums,0,nums.size()-1,scores)>=target;

    }
    int maxScore(vector& nums,int start,int end,vector> scores)
    {
        if(start>end)
            return 0;
        if(start==end)
            return nums[start];
        if(scores[start][end])
            return scores[start][end];
        int res1=nums[start]+min(maxScore(nums,start+2,end,scores),maxScore(nums,start+1,end-1,scores));
        int res2=nums[end]+min(maxScore(nums,start,end-2,scores),maxScore(nums,start+1,end-1,scores));
        scores[start][end]=max(res1,res2);
        return scores[start][end];
    }
};

方法二（超时）：

class Solution{
public:
    bool PredictTheWinner(vector& nums)
    {
        //vector[start][end]代表nums的头索引为start,尾索引为end时player1是否比player2大，即是否大于等于0
        vector> scores(nums.size(),vector(nums.size(),INT_MAX));
        int res=maxScore(nums,0,nums.size()-1,scores);
        return res>=0;

    }
    int maxScore(vector& nums,int start,int end,vector> scores)
    {
        if(scores[start][end]==INT_MAX)
            scores[start][end]=start==end?nums[start]:max(nums[start]-maxScore(nums,start+1,end,scores),
            nums[end]-maxScore(nums,start,end-1,scores));
        return scores[start][end];
    }
};

方法三(方法二的非递归版)：

class Solution{
public:
    bool PredictTheWinner(vector& nums) {
        //vector[start][end]代表nums的头索引为start,尾索引为end时一个player是否比另一个player大，即是否大于等于0
        //vector[i][j]又可以解释为以某一个选手开始，最终比另一个选手的最多高出的值。
        vector> scores(nums.size(), vector(nums.size(), 0));
        for(int i=0;i=0;i--)
        {
            for(int j=i+1;j=0;
    }
};