[LeetCode] 486. Predict the Winner 预测赢家

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.

这道题给了一个小游戏，有一个数组，两个玩家轮流取数，说明了只能从开头或结尾取，问我们第一个玩家能赢吗。这道题博主想到了应该是用 Minimax 来做，由于之前有过一道这样的题 Guess Number Higher or Lower II，所以依稀记得应该要用递归的方法，而且当前玩家赢返回 true 的条件就是递归调用下一个玩家输返回 false。这里需要一个变量来标记当前是第几个玩家，还需要两个变量来分别记录两个玩家的当前数字和，在递归函数里面，如果当前数组为空了，直接比较两个玩家的当前得分即可，如果数组中只有一个数字了，根据玩家标识来将这个数字加给某个玩家并进行比较总得分。如果数组有多个数字，分别生成两个新数组，一个是去掉首元素，一个是去掉尾元素，然后根据玩家标识分别调用不同的递归，只要下一个玩家两种情况中任意一种返回 false 了，那么当前玩家就可以赢了，参见代码如下：

解法一：

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        return canWin(nums, 0, 0, 1);
    }
    bool canWin(vector<int> nums, int sum1, int sum2, int player) {
        if (nums.empty()) return sum1 >= sum2;
        if (nums.size() == 1) {
            if (player == 1) return sum1 + nums[0] >= sum2;
            else if (player == 2) return sum2 + nums[0] > sum1;
        }
        vector<int> va = vector<int>(nums.begin() + 1, nums.end());
        vector<int> vb = vector<int>(nums.begin(), nums.end() - 1);
        if (player == 1) {
            return !canWin(va, sum1 + nums[0], sum2, 2) || !canWin(vb, sum1 + nums.back(), sum2, 2);
        } else if (player == 2) {
            return !canWin(va, sum1, sum2 + nums[0], 1) || !canWin(vb, sum1, sum2 + nums.back(), 1);
        }
    }
};

我们还可以使用 DP 加 Minimax 的方法来做，先来看递归的写法，十分的简洁。DP 数组的作用是保存中间结果，再次遇到相同情况时直接返回不用再次计算，提高了运算效率：

解法二：

class Solution {
public:
    bool PredictTheWinner(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> dp(n, vector<int>(n, -1));
        return canWin(nums, 0, n - 1, dp) >= 0;
    }
    int canWin(vector<int>& nums, int s, int e, vector<vector<int>>& dp) {
        if (dp[s][e] == -1) {
            dp[s][e] = (s == e) ? nums[s] : max(nums[s] - canWin(nums, s + 1, e, dp), nums[e] - canWin(nums, s, e - 1, dp));
        }
        return dp[s][e];
    }
};

下面这种方法是 DP 加 Minimax 的迭代写法，要注意的是 DP 的更新顺序，跟以往不太一样，这种更新方法是按区间来更新的，感觉之前好像没有遇到过这种更新的方法，还蛮特别的：

解法三：

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