Update 60. n 个骰子的点数.md

题目链接改为力扣中的对应题目,题目描述和代码按该题的接口调整。
This commit is contained in:
CyC2018
2026-10-28 13:45:07 +08:00
parent 3b095c8ec0
commit e68818f002
+16 -16
View File
@@ -2,7 +2,7 @@
## 题目链接
[Lintcode](https://www.lintcode.com/en/problem/dices-sum/)
[Leetcode](https://leetcode.cn/problems/nge-tou-zi-de-dian-shu-lcof/)
## 题目描述
@@ -19,23 +19,23 @@
空间复杂度:O(N<sup>2</sup>)
```java
public List<Map.Entry<Integer, Double>> dicesSum(int n) {
public double[] statisticsProbability(int num) {
final int face = 6;
final int pointNum = face * n;
long[][] dp = new long[n + 1][pointNum + 1];
final int pointNum = face * num;
long[][] dp = new long[num + 1][pointNum + 1];
for (int i = 1; i <= face; i++)
dp[1][i] = 1;
for (int i = 2; i <= n; i++)
for (int i = 2; i <= num; i++)
for (int j = i; j <= pointNum; j++) /* 使用 i 个骰子最小点数为 i */
for (int k = 1; k <= face && k <= j; k++)
dp[i][j] += dp[i - 1][j - k];
final double totalNum = Math.pow(6, n);
List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
for (int i = n; i <= pointNum; i++)
ret.add(new AbstractMap.SimpleEntry<>(i, dp[n][i] / totalNum));
final double totalNum = Math.pow(6, num);
double[] ret = new double[pointNum - num + 1];
for (int i = num; i <= pointNum; i++)
ret[i - num] = dp[num][i] / totalNum;
return ret;
}
@@ -46,16 +46,16 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n) {
空间复杂度:O(N)
```java
public List<Map.Entry<Integer, Double>> dicesSum(int n) {
public double[] statisticsProbability(int num) {
final int face = 6;
final int pointNum = face * n;
final int pointNum = face * num;
long[][] dp = new long[2][pointNum + 1];
for (int i = 1; i <= face; i++)
dp[0][i] = 1;
int flag = 1; /* 旋转标记 */
for (int i = 2; i <= n; i++, flag = 1 - flag) {
for (int i = 2; i <= num; i++, flag = 1 - flag) {
for (int j = 0; j <= pointNum; j++)
dp[flag][j] = 0; /* 旋转数组清零 */
@@ -64,10 +64,10 @@ public List<Map.Entry<Integer, Double>> dicesSum(int n) {
dp[flag][j] += dp[1 - flag][j - k];
}
final double totalNum = Math.pow(6, n);
List<Map.Entry<Integer, Double>> ret = new ArrayList<>();
for (int i = n; i <= pointNum; i++)
ret.add(new AbstractMap.SimpleEntry<>(i, dp[1 - flag][i] / totalNum));
final double totalNum = Math.pow(6, num);
double[] ret = new double[pointNum - num + 1];
for (int i = num; i <= pointNum; i++)
ret[i - num] = dp[1 - flag][i] / totalNum;
return ret;
}