一、原题描述
n passengers board an airplane with exactly n seats. The first passenger has lost the ticket and picks a seat randomly. But after that, the rest of passengers will:
- Take their own seat if it is still available,
- Pick other seats randomly when they find their seat occupied
What is the probability that the n-th person can get his own seat?
Example 1:
Input: n = 1
Output: 1.00000
Explanation: The first person can only get the first seat.
Example 2:
Input: n = 2
Output: 0.50000
Explanation: The second person has a probability of 0.5 to get the second seat (when first person gets the first seat）
二、简要翻译
n个用户依次登机就坐。第一个用户丢失了机票，将会随机选取一个座位，之后的乘客优先坐自己的座位，如果自己座位被占了则随机找寻一个座位就坐，求问第n个用户得到自己座位的概率。
三、解答与分析
1、代码
public double nthPersonGetsNthSeat(int n) {
if (n == 1) {
return 1.0;
} else {
return 0.5;
}
}
2、分析与证明
- 假设有n个用户，本问题的答案为 f(n)。
- 如果第一个人随机到了自己的位置，那么后面的人一定按自己机票座位号入座。
- 如果第一个人随机到了第n个人的位置，那么第 n 个人得到自己座位的概率为0。
- 如果第一个人随机到了第2个人的位置，那么第 n 个人得到自己座位的概率为f(n-1)。
- 依次类推可得 f(n) = (1 + f(n-1) + f(n-2) + ... + f(2) + 0) / n ;
- 假设当 1< i <= k 时 f(i) = 1/2 , 容易证明f(k+1) = 1/2; 所以f(n) 在n > 1的时候恒等于 1/2 .
论证过程的代码实现如下
public double nthPersonGetsNthSeat(int n) {
if(n==1) return 1.0;
double[] dp = new double[n];
double sum = 0;
for (int i = 1; i < n; i++) {
dp[i] = (1 + sum) / (i + 1);
sum += dp[i];
}
return dp[n - 1];
}