LeetCode 837. New 21 Game

原题链接在这里：https://leetcode.com/problems/new-21-game/

题目：

Alice plays the following game, loosely based on the card game "21".

Alice starts with 0 points, and draws numbers while she has less than K points.  During each draw, she gains an integer number of points randomly from the range [1, W], where W is an integer.  Each draw is independent and the outcomes have equal probabilities.

Alice stops drawing numbers when she gets K or more points.  What is the probability that she has N or less points?

Example 1:

Input: N = 10, K = 1, W = 10
Output: 1.00000
Explanation:  Alice gets a single card, then stops.

Example 2:

Input: N = 6, K = 1, W = 10
Output: 0.60000
Explanation:  Alice gets a single card, then stops.
In 6 out of W = 10 possibilities, she is at or below N = 6 points.

Example 3:

Input: N = 21, K = 17, W = 10
Output: 0.73278

Note:

1. 0 <= K <= N <= 10000
2. 1 <= W <= 10000
3. Answers will be accepted as correct if they are within 10^-5 of the correct answer.
4. The judging time limit has been reduced for this question.

题解：

When the draws sum up to K, it stops, calculate the possibility K<=sum<=N.

Think about one step earlier, sum = K-1, game is not ended and draw largest card W. K-1+W is the maximum sum could get when game is ended. If it is <= N, then for sure the possiblity when games end ans sum <= N is 1.

Because the maximum is still <= 1.

Otherwise calculate the possibility sum between K and N. 

Let dp[i] denotes the possibility of that when game ends sum up to i.

i is a number could be got equally from i - m and draws value m card.

Then dp[i] should be sum of dp[i-W] + dp[i-W+1] + ... + dp[i-1], devided by W.

We only need to care about previous W value sum, accumlate winSum, reduce the possibility out of range.

Time Complexity: O(N).

Space: O(N).

AC Java:   

 1 class Solution {
 2     public double new21Game(int N, int K, int W) {
 3         if(K == 0 || K-1+W <= N){
 4             return 1;
 5         }
 6         
 7         if(K > N){
 8             return 0;
 9         }
10         
11         double [] dp = new double[N+1];
12         dp[0] = 1.0;
13         double winSum = 1;
14         
15         double res = 0.0;
16         for(int i = 1; i<=N; i++){
17             dp[i] = winSum/W;
18             
19             if(i<K){
20                 winSum += dp[i];
21             }else{
22                 res += dp[i];
23             }
24             
25             if(i >= W){
26                 winSum -= dp[i-W];
27             }
28         }
29         
30         return res;
31     }
32 }

类似Climbing Stairs.

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/11490953.html