LintCode 396: Coins in a Line III (博弈类DP经典题)

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本文探讨了在两人轮流取硬币游戏中，如何通过动态规划算法预测先手玩家是否能获胜。通过构建状态转移方程，我们分析了在不同硬币分布下，玩家的最佳策略选择，最终得出先手玩家的胜率。

Coins in a Line III

There are n coins in a line. Two players take turns to take a coin from one of the ends of the line until there are no more coins left. The player with the larger amount of money wins.

Could you please decide the first player will win or lose?

Example
Given array A = [3,2,2], return true.

Given array A = [1,2,4], return true.

Given array A = [1,20,4], return false.

Challenge
Follow Up Question:

If n is even. Is there any hacky algorithm that can decide whether first player will win or lose in O(1) memory and O(n) time?

解法1：DP。
思路：还是把两个选手当成一个人，每次面对a[i…j]选最优解。
这里dp[i][j]是当前选手对a[i…j]所选的coins与对手将要选的coins的最大差值。
转移方程：
f[i][j] = max{a[i] - f[i+1][j], a[j] - f[i][j-1]}
注意：

初始条件：dp[i][i] = values[i]。因为只有1个coin，肯定是全选。
dp计算顺序为：
dp[0][0], dp[1][1], dp[2][2], …, dp[n-1][n-1] //len = 0, 1 coin
dp[0][1], …, dp[n-2][n-1] //len = 1, 2 coins
…
dp[0][n-1] //len = n - 1, n coins
代码如下：

class Solution {
public:
    /**
     * @param values: a vector of integers
     * @return: a boolean which equals to true if the first player will win
     */
    bool firstWillWin(vector<int> &values) {
        int n = values.size();
        if (n <= 1) return true;
        vector<vector<int>> dp(n, vector<int>(n, 0));
        
        for (int i = 0; i < n; ++i) {
            dp[i][i] = values[i];
        }
        
        for (int len = 1; len < n; ++len) {
            for (int i = 0; i + len < n; ++i) {
                int j = i + len;
                dp[i][j] =  max(values[i] - dp[i + 1][j], values[j] - dp[i][j - 1]);
            }
        }
        return dp[0][n - 1] >= 0;
    }
};