剑指 Offer 60. n个骰子的点数

class Solution {
    public double[] dicesProbability(int n) {
        // 定义 dp[i][j] 为 i 个骰子时点数之和为 j 时的概率
        double[][] dp = new double[n + 1][6 * n + 1];
        dp[1][1] = dp[1][2] = dp[1][3] = dp[1][4] = dp[1][5] = dp[1][6] = 1.0 / 6;

        for (int i = 2; i <= n; i++) {
            // 注意 j 的取值范围为 [i, 6 * i]
            for (int j = i; j <= 6 * i; j++) {
                for (int k = 1; k <= 6; k++) {
                    if (j - k > 0) {
                        dp[i][j] += dp[i - 1][j - k] * 1 / 6; // 上一个骰子点数为 j - k 时贡献的概率
                    }
                }
            }
        }

        double[] res = new double[6 * n - n + 1];
        for (int i = n; i <= 6 * n; i++) {
            res[i - n] = dp[n][i];
        }
        return res;
    }
}
Reference
