剑指 Offer 60. n个骰子的点数
本问题对应的 leetcode 原文链接：剑指 Offer 60. n个骰子的点数
问题描述
把n个骰子扔在地上，所有骰子朝上一面的点数之和为s。输入n，打印出s的所有可能的值出现的概率。
你需要用一个浮点数数组返回答案，其中第 i 个元素代表这 n 个骰子所能掷出的点数集合中第 i 小的那个的概率。
示例 1:
输入: 1
输出: [0.16667,0.16667,0.16667,0.16667,0.16667,0.16667]
示例 2:
输入: 2
输出: [0.02778,0.05556,0.08333,0.11111,0.13889,0.16667,0.13889,0.11111,0.08333,0.05556,0.02778]
限制：
- 1 <= n <= 11
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public double[] dicesProbability(int n) {
int[][] dp = new int[n+1][6*n+1];
for(int i = 1; i <= 6; i++){
dp[1][i] = 1;
}
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) break;
dp[i][j] += dp[i-1][j-k];
}
}
}
double[] res = new double[5*n + 1];
int index = 0;
double sum = Math.pow(6, n);
for(int i = n; i <= 6 * n; i++){
res[index++] = dp[n][i] / sum;
}
return res;
}
}
时间复杂度：O(n^2)
空间复杂度：O(n)
Python
class Solution(object):
def dicesProbability(self, n):
"""
:type n: int
:rtype: List[float]
"""
dp = [[0]*(6*n+1) for _ in range(n+1)]
for i in range(1, 7):
dp[1][i] = 1
for i in range(2, n+1):
for j in range(i, 6*i+1):
for k in range(1, 7):
if j < k:
break
dp[i][j] += dp[i-1][j-k]
res = [0]*(5*n+1)
index = 0
s = pow(6, n)
for i in range(n, 6*n+1):
res[index] = float(dp[n][i]) / s
index += 1
return res
C++
class Solution {
public:
vector<double> dicesProbability(int n) {
vector<vector<int>> dp(n+1, vector<int>(6*n+1));
for (int i = 1; i <= 6; i++) {
dp[1][i] = 1;
}
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) break;
dp[i][j] += dp[i-1][j-k];
}
}
}
vector<double> res(5*n + 1);
int index = 0;
double sum = pow(6, n);
for (int i = n; i <= 6 * n; i++) {
res[index++] = static_cast<double>(dp[n][i]) / sum;
}
return res;
}
};
Go
func dicesProbability(n int) []float64 {
dp := make([][]int, n+1)
for i := range dp {
dp[i] = make([]int, 6*n+1)
}
for i := 1; i <= 6; i++ {
dp[1][i] = 1
}
for i := 2; i <= n; i++ {
for j := i; j <= 6*i; j++ {
for k := 1; k <= 6; k++ {
if j < k {
break
}
dp[i][j] += dp[i-1][j-k]
}
}
}
res := make([]float64, 0, 5*n+1)
s := math.Pow(6, float64(n))
for i := n; i <= 6*n; i++ {
res = append(res, float64(dp[n][i])/s)
}
return res
}
JS
/**
* @param {number} n
* @return {number[]}
*/
var dicesProbability = function(n) {
const dp = new Array(n+1).fill().map(() => new Array(6*n+1).fill(0));
for (let i = 1; i <= 6; i++) {
dp[1][i] = 1;
}
for (let i = 2; i <= n; i++) {
for (let j = i; j <= 6*i; j++) {
for (let k = 1; k <= 6; k++) {
if (j < k) break;
dp[i][j] += dp[i-1][j-k];
}
}
}
const res = new Array(5*n + 1).fill(0);
let index = 0;
const sum = Math.pow(6, n);
for (let i = n; i <= 6 * n; i++) {
res[index++] = dp[n][i] / sum;
}
return res;
};