LeetCode 837. New 21 Game 滑动窗口逆向动态规划

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本文介绍了一种基于卡牌游戏“21”的概率计算问题，Alice的目标是在获得K点或更多点时停止抽牌，每次抽牌从1到W的范围内随机获取整数点数。文章提供了两种算法解决方案，一种是滑动窗口法，另一种是动态规划法，用于计算Alice在N点或更少的情况下获胜的概率。

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:

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

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直接搞N*W的复杂度肯定会超时，这题要优化到O(N)的复杂度，有两种思路：

思路一，滑动窗口：，位置i的概率是有[i-1,i-W]跳过去的，因此是1/W*sum([i-1,i-W]),注意[i-1,i-W]这个区间窗口滑动滑动就滑不下去了，因此过滤掉边界即可：

class Solution:
    def new21Game(self, N, K, W):
        if K == 0 or N >= K + W: return 1
        dp = [1.0] + [0.0] * N
        sliding = 1.0
        for i in range(1, N + 1):
            dp[i] = sliding / W
            if i < K: sliding += dp[i]
            if i - W >= 0: sliding -= dp[i - W]
            #print("sliding={0}".format(sliding))
        return sum(dp[K:])

思路二，动态规划：如果起点位于闭区间[K,N]，那么 the probability that she has N or less points is 1, i.e., f(x)=1 when K<=x<=N, f(x)=0 when x>N. 其中f(x)表示起点从x开始，the probability that she has N or less points，最后求f(0)就可以。f(x)=1/W*(f(x+1)+...f(x+W))，而f(x-1)=1/W*(f(x)+...f(x+W-1))，这样根据f(x)和f(x-1)就可以得到O(1)的递归公式。那么codes是

class Solution(object):
    def new21Game(self, N, K, W):
        dp = [0.0] * (N + W + 1)
        # dp[x] = the answer when Alice has x points
        for k in xrange(K, N + 1):
            dp[k] = 1.0

        S = min(N - K + 1, W)
        # S = dp[k+1] + dp[k+2] + ... + dp[k+W]
        for k in xrange(K - 1, -1, -1):
            dp[k] = S / float(W)
            S += dp[k] - dp[k + W]

        return dp[0]