题目
解
采用了最高浏览量的答案——动态规划
代码一,没有进行空间优化
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;
}
};
代码二——采用了空间优化
class Solution {
public:
vector<double> twoSum(int n) {
int dp[70];//存状态
memset(dp, 0, sizeof(dp));//存储点数之和出现的次数,并初始化为0
for (int i = 1; i <= 6; i++)
dp[i] = 1;//初始化一个骰子的情形
for (int i = 2; i <= n; i++)//从2个骰子到n个骰子的情况
{
for (int j = 6*i; j >= i; j--)//点数之和为i-6i
{
dp[j]=0;//重置本轮该点数信息
for (int cur = 1; cur <= 6; cur++)//从后往前,点数j的次数由前六次之和构成
{
if(j-cur<i-1) break;//状态递推中,加和只加到上一轮的首项
dp[j] += dp[j - cur];//即状态转移
}
}
}
int all = pow(6, n);
vector<double> ans;
for (int i = n; i <= 6 * n; i++)//n个骰子点数之和的所有可能性
{
ans.push_back(1.0*dp[i] / all);
}
return ans;
}
};
提交结果
优秀!