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.

不管是谁拿， 都希望自己拿到的总和是最大的， 

/* 两人交替从数组两头取数 结果和最大的取胜
 * 这是minimax 问题 即假设对手每次按最优策略。博弈论
 * 关键是有两个人 假设有个函数partition 结果是甲目前总和-乙目前总和 结果>0显然甲胜利
 * parit(int begin, end, dp[][]) dp[beg][end] 记录partition在begin-end间的值 相当于递归到终点时 向外返回时不断填充dp的值(递归写法)
 * */
class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        int len=nums.size();
        vector<vector<int>> dp(len, vector<int>(len, -1));
        return divide(nums, 0, len-1, dp) >=0;
    }
    // 边界条件数组拿空了
    int divide(const vector<int> &nums, int begin, int end, vector<vector<int>> &dp){
        if(dp[begin][end] == -1){// 防止重复进入
            if(begin == end)    dp[begin][end] = nums[begin];
            else{
                dp[begin][end] = max(nums[begin] - divide(nums, begin+1, end, dp), nums[end]-divide(nums, begin, end-1, dp));
            }
        }
        return dp[begin][end];
    }
};

非递归写法（注意更新顺序）

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        int len=nums.size();
        vector<vector<int>> dp(len, vector<int>(len, 0));
        // dp[i][j] 为取i-j双方博弈互最优结果 即j-i = 2k时 是甲最优结果 j-i=2k+1 时是乙最优结果 两人轮流来 共用一个dp
        // 非递归形式  dp[i+1][j] dp[i][j-1] 边界条件i==j的所有位置
        for(int i=0;i<len;i++){
            dp[i][i] = nums[i];
        }
        // 以下循环顺序是错的 以 1 2 3 为例 循环更新顺序为 1-2 1-3 2-3 最后更新的是2-3 而实际要求的为1-3
        // 更新顺序应为小范围区间扩展到大范围区间即 2-3 1-2 1-3
//        for(int i=0;i<len-1;i++){
//            for(int j=i+1;j<len;j++){
//                dp[i][j] = max(nums[i]-dp[i+1][j], nums[j]-dp[i][j-1]);
//            }
//        }
        for(int i=len-2;i>=0;i--){
            for(int j=i+1;j<len;j++){
                dp[i][j] = max(nums[i]-dp[i+1][j], nums[j]-dp[i][j-1]);
            }
        }
        return dp[0][len-1]>=0;
    }

};

之前字符串匹配时， dp[][]更新顺序是i=0->len, j=0->len; 是因为dp[i][j] 表示s[0：i]和p[0:j]是否匹配上，这里dp[i][j]就是表示num[i:j]之间最优决策得到的结果