leetcode-983-最低票价

题目描述：

 

 

 

 方法一：O(W)  O(W) W =365

class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        dayset = set(days)
        durations = [1,7,30]

        @lru_cache(None)
        def dp(i):
            if i > 365:
                return 0
            elif i in dayset:
                return min(dp(i + d) + c for c,d in zip(costs,durations))
            else:
                return dp(i+1)

        return dp(1)

方法二：O(N) O(N)

class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        N = len(days)
        durations = [1,7,30]

        @lru_cache(None)
        def dp(i)
        if i >= N:
            return 0
        ans = 10**9
        j = i
        for c,d in zip(costs,durations):
            while j < N and days[j] < days[i] + d:
                j += 1
            ans = min(ans,dp(j) + c)
        return ans

    return dp(0)

 另：

class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        dp = [0 for _ in range(days[-1] + 1)]  # dp数组，每个元素代表到当前天数最少钱数，为下标方便对应，多加一个 0 位置
        days_idx = 0  # 设定一个days指标，标记应该处理 days 数组中哪一个元素
        for i in range(1, len(dp)):
            if i != days[days_idx]:  # 若当前天数不是待处理天数，则其花费费用和前一天相同
                dp[i] = dp[i - 1]
            else:
                # 若 i 走到了待处理天数，则从三种方式中选一个最小的
                dp[i] = min(dp[max(0, i - 1)] + costs[0],
                            dp[max(0, i - 7)] + costs[1],
                            dp[max(0, i - 30)] + costs[2])
                days_idx += 1
        return dp[-1]  # 返回最后一天对应的费用即可

作者：LotusPanda
链接：https://leetcode-cn.com/problems/minimum-cost-for-tickets/solution/xiong-mao-shua-ti-python3-dong-tai-gui-hua-yi-do-2/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。