[leetcode] 486. Predict the Winner (medium)

原题

思路：
解法一：
转换比较拿取分数多少的思路，改为考虑 player拿的分数为正，把Player2拿的视为负，加上所有分数，如果最后结果大于0则Player1赢。
思考得出递归表达式：
max(nums[beg] - player2(beg + 1, end), nums[end] - player2(beg, end + 1))
此解法效率很差 104ms，beats 5.32%

class Solution {
 public:
  bool PredictTheWinner(vector<int> &nums) {
    return helper(0, nums.size() - 1, nums) >= 0;
  }
  int helper(int beg, int end, vector<int> &nums) {
    if (beg >= end) return nums[beg];
    return max(nums[beg] - helper(beg + 1, end, nums),
               nums[end] - helper(beg, end - 1, nums));
  }
};

解法二：
discussing里看到的利用dp结合MiniMax的优解

class Solution {
public:
    
    int findwin(vector<int>&v, int left, int right, vector<vector<int>>& dp){
        if(left > right)
            return 0;
        if(dp[left][right] != -1)
            return dp[left][right];
        int pos1 = v[left] + min(findwin(v, left + 2, right, dp), findwin(v, left+1, right-1, dp));
        int pos2 = v[right] + min(findwin(v, left+1, right-1, dp), findwin(v, left, right-2, dp));
        return dp[left][right] = max(pos1, pos2);
    }
    
    bool PredictTheWinner(vector<int>& v) {
        if(v.size() == 0)
            return false;
        if(v.size() == 1)
            return true;
        vector<vector<int>> dp(v.size(), vector<int>(v.size(), -1));
        int left = 0;
        int right = v.size()-1;
        int player1 = findwin(v, left, right, dp);
        int sum = 0;
        for(int i=0; i<v.size(); i++)
            sum += v[i];
        return player1 >= sum - player1;
    }
};