365天挑战LeetCode1000题——Day 106 石子游戏 I - V

石子游戏 I

代码实现

class Solution {
public:
    bool stoneGame(vector<int>& piles) {
        int n = piles.size();
        vector<vector<int>> dp(n, vector<int>(n));
        for (int i = 0; i < n; i  ) {
            dp[i][i] = piles[i];
        }
        for (int i = n - 2; i >= 0; i--) {
            for (int j = i   1; j < n; j  ) {
                dp[i][j] = max(piles[i] - dp[i   1][j], piles[j] - dp[i][j - 1]);
            }
        }
        return dp[0][n - 1];
    }
};

石子游戏 II

代码实现

class Solution {
public:
    int stoneGameII(vector<int>& piles) {
        int n = piles.size();
        vector<int> sum(n   1, 0);
        for (int i = 1; i <= n; i  ) {
            sum[i] = sum[i - 1]   piles[i - 1];
        }
        vector<vector<int>> dp(n   1, vector<int>(n   1));
        for (int i = 0; i <= n; i  ) {
            for (int j = i; j <= n; j  ) {
                dp[i][j] = sum[n] - sum[n - i];
            }
        }
        for (int i = 1; i <= n; i  ) {
            for (int j = 1; j <= n; j  ) {
                for (int k = 1; k <= 2 * j && k <= i;   k) {
                    dp[i][j] = max(dp[i][j], sum[n] - sum[n - i] - dp[i - k][min(max(k, j), n)]);
                }
            }
        }
        
        return dp[n][1];
    }
};

石子游戏 III

代码实现

class Solution {
public:
    string stoneGameIII(vector<int>& stoneValue) {
        int n = stoneValue.size();
        if (n == 1) return stoneValue[0] > 0 ? "Alice" : (stoneValue[0] == 0 ? "Tie" : "Bob");
        vector<int> sum(n   1);
        for (int i = 1; i <= n; i  ) {
            sum[i] = sum[i - 1]   stoneValue[i - 1];
        }
        vector<vector<int>> dp(n   1, vector<int>(4, INT_MIN));
        for (int i = 0; i <= 2; i  ) {
            for (int j = 1; j <= 3; j  ) {
                dp[i][j] = sum[min(n - i   j, n)] - sum[n - i];
            }
        }
        // 还剩下i个石头
        for (int i = 3; i <= n; i  ) {
            // 选择走j步
            for (int j = 1; j <= 3; j  ) {
                // 选取最大值
                // cout << i << " " << j << " " << k << endl;
                dp[i][j] = max(dp[i][j], 
                sum[n] - sum[n - i] - max(dp[i - j][1], max(dp[i - j][2], dp[i - j][3]))); 
            }
        }
        int ans = max(dp[n][1], max(dp[n][2], dp[n][3]));
        if (ans * 2 > sum[n]) return "Alice";
        else if (ans * 2 == sum[n]) return "Tie";
        else return "Bob";
    }
};

石子游戏 IV

代码实现

class Solution {
public:
    bool winnerSquareGame(int n) {
        // 定义状态dp[i]，表示还剩i个石子时的胜利情况
        vector<bool> dp(n   1, false);
        dp[0] = false;
        for (int i = 1; i * i <= n; i  ) {
            dp[i * i] = true;
        }
        for (int i = 2; i <= n; i  ) {
            for (int j = 1; j * j <= i; j  ) {
                dp[i] = dp[i] || !dp[i - j * j];
            }
        }
        return dp[n];
    }
};

石子游戏 V

代码实现

class Solution {
public:
    int stoneGameV(vector<int>& stoneValue) {
        // dp[i][j]，当序列为[i, j]时，Alice能拿到的最大分数
        int n = stoneValue.size();
        vector<int> sum(n   1);
        for (int i = 1; i <= n; i  ) {
            sum[i] = sum[i - 1]   stoneValue[i - 1];
        }
        int left, total, i, j, k;
        int dp[501][501];
        memset(dp, 0, sizeof(dp));
        // vector<vector<int>> dp(n   1, vector<int>(n   1));
        for (i = n - 1; i >= 1; i--) {
            for (j = i   1; j <= n; j  ) {
                total = sum[j] - sum[i - 1];
                for (k = i; k < j; k  ) {
                    left = sum[k] - sum[i - 1];
                    if (left * 2 < total) {
                        dp[i][j] = max(dp[i][j], left   dp[i][k]);
                    }
                    else if (left * 2 == total) {
                        dp[i][j] = max(max(dp[i][j], left   dp[i][k]),
                        left   dp[k   1][j]);
                    }
                    else {
                        dp[i][j] = max(dp[i][j], total - left   dp[k   1][j]);
                    }
                }
                //cout <<"dp["<<i<<"]"<<"["<<j<<"]:"<<dp[i][j] << endl;
            }
        }
        return dp[1][n];
    }
};

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