题目描述
扔 n 个骰子,向上面的数字之和为 S。
给定 n,请列出所有可能的 S 值及其相应的概率。
样例1
输入:n = 1
输出:[[1, 0.17], [2, 0.17], [3, 0.17], [4, 0.17], [5, 0.17], [6, 0.17]]
解释:掷一次骰子,向上的数字和可能为1,2,3,4,5,6,出现的概率均为 0.17。
样例2
输入:n = 2
输出:[[2,0.03],[3,0.06],[4,0.08],[5,0.11],[6,0.14],[7,0.17],[8,0.14],[9,0.11],[10,0.08],[11,0.06],[12,0.03]]
解释:掷两次骰子,向上的数字和可能在[2,12],出现的概率是不同的。
java题解
public class Solution {
/**
* @param n an integer
* @return a list of Map.Entry<sum, probability>
*/
public List<Map.Entry<Integer, Double>> dicesSum(int n) {
// Write your code here
// Ps. new AbstractMap.SimpleEntry<Integer, Double>(sum, pro)
// to create the pair.
List<Map.Entry<Integer, Double>> results =
new ArrayList<Map.Entry<Integer, Double>>();
double[][] f = new double[n + 1][6 * n + 1];
for (int i = 1; i <= 6; ++i)
f[1][i] = 1.0 / 6;
for (int i = 2; i <= n; ++i)
for (int j = i; j <= 6 * n; ++j) {
for (int k = 1; k <= 6; ++k)
if (j > k)
f[i][j] += f[i - 1][j - k];
f[i][j] /= 6.0;
}
for (int i = n; i <= 6 * n; ++i)
results.add(new AbstractMap.SimpleEntry<Integer, Double>(i, f[n][i]));
return results;
}
}
C++题解
class Solution {
public:
/**
* @param n an integer
* @return a list of pair<sum, probability>
*/
vector<pair<int, double>> dicesSum(int n) {
// Write your code here
vector<pair<int, double>> results;
vector<vector<double>> f(n + 1, vector<double>(6 * n + 1));
for (int i = 1; i <= 6; ++i) f[1][i] = 1.0 / 6;
for (int i = 2; i <= n; ++i)
for (int j = i; j <= 6 * i; ++j) {
for (int k = 1; k <= 6; ++k)
if (j > k)
f[i][j] += f[i - 1][j - k];
f[i][j] /= 6.0;
}
for (int i = n; i <= 6 * n; ++i)
results.push_back(make_pair(i, f[n][i]));
return results;
}
};
python题解
class Solution:
# @param {int} n an integer
# @return {tuple[]} a list of tuple(sum, probability)
def dicesSum(self, n):
# Write your code here
results = []
f = [[0 for j in xrange(6 * n + 1)] for i in xrange(n + 1)]
for i in xrange(1, 7):
f[1][i] = 1.0 / 6.0
for i in xrange(2, n + 1):
for j in xrange(i, 6 * n + 1):
for k in xrange(1, 7):
if j > k:
f[i][j] += f[i - 1][j - k]
f[i][j] /= 6.0
for i in xrange(n, 6 * n + 1):
results.append((i, f[n][i]))
return results