一.问题
-
https://leetcode-cn.com/problems/nge-tou-zi-de-dian-shu-lcof/:面试题60:n个骰子的点数
二.解法
-
动态规划:用 dp[n][j] 来表示n个骰子的点数 j 出现的次数。
-
状态转移方程:表示n个骰子产生的点数 j 出现的次数等于n-1个骰子产生 j - 1, j - 2, j - 3, j - 4, j - 5, j - 6 次数之和
-
基本步骤:1.先设立参数以及参数对应的意义,2.写出动态转移方程,3.设置初始状态值
1.代码一:
class Solution {
public:
vector<double> twoSum(int n) {
int dp[15][70];
memset(dp, 0, sizeof(dp));
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 cur = 1; cur <= 6; cur ++) {
if (j - cur <= 0) {
break;
}
dp[i][j] += dp[i-1][j-cur];
}
}
}
int all = pow(6, n);
vector<double> ret;
for (int i = n; i <= 6 * n; i ++) {
ret.push_back(dp[n][i] * 1.0 / all);
}
return ret;
}
};
2.代码二
-
优化二维数组变为一维数组:方法是从后向前修改值,保证前一维的值不变来得到后一维的值
class Solution {
public:
vector<double> twoSum(int n) {
int dp[70];
memset(dp, 0, sizeof(dp));
for (int i = 1; i <= 6; i ++) {
dp[i] = 1;
}
for (int i = 2; i <= n; i ++) {
for (int j = 6*i; j >= i; j --) {
dp[j] = 0;
for (int cur = 1; cur <= 6; cur ++) {
if (j - cur < i-1) {
break;
}
dp[j] += dp[j-cur];
}
}
}
int all = pow(6, n);
vector<double> ret;
for (int i = n; i <= 6 * n; i ++) {
ret.push_back(dp[i] * 1.0 / all);
}
return ret;
}
};
三.思路和代码来自:
作者:huwt
链接:https://leetcode-cn.com/problems/nge-tou-zi-de-dian-shu-lcof/solution/nge-tou-zi-de-dian-shu-dong-tai-gui-hua-ji-qi-yo-3/