Leetcode 1696. Jump Game VI (python)

题目

解法1:暴力dp（TLE）

class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        if not nums:
            return 0
        
        dp = [float(-inf)]*len(nums)
        dp[0] = nums[0]
        
        for i in range(len(nums)):
            for j in range(max(i-k,0),i):
                dp[i] = max(dp[j]+nums[i],dp[i])
        
        #print(dp)
        return dp[-1]

解法2：dp+heap

可以发现这边dp的时候，只需要找到之前能够跳到当前这步中的最大值即可，那么可以用heap来维护，把index也放入dp来帮助判断是否能从这个状态转移。复杂度n(logn)

class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        if not nums:
            return 0
        
        dp = [float(-inf)]*len(nums)
        dp[0] = nums[0]
        max_heap = []
        heapq.heappush(max_heap,(-nums[0],0))
        
        for i in range(1,len(nums)):
            while max_heap[0][1]<i-k:
                heapq.heappop(max_heap)
            dp[i] = nums[i] - max_heap[0][0]
            heapq.heappush(max_heap,(-dp[i],i))
            
        
        #print(dp)
        return dp[-1]

解法3：dp+sliding window

仔细观察可以发现这边的步数k实际就像是一个窗口，从这个角度出发就可以结合sliding window maximum的解法，因为sliding window的解法，每个元素只会入队列一次和出队列一次，所以总体解法就是O(n)的。关于sliding window maximum可以看这边

class Solution:
    def maxResult(self, nums: List[int], k: int) -> int:
        if not nums:
            return 0
        
        dp = [float(-inf)]*len(nums)
        dp[0] = nums[0]
        dq = collections.deque()
        dq.append(0)
        
        for i in range(1,len(nums)):
            # similar to liding window maximum
            while dq and dq[0]<i-k:
                dq.popleft()
            dp[i] = nums[i] + dp[dq[0]]
            while dq and dp[dq[-1]]<=dp[i]:
                dq.pop()
            dq.append(i)
              
        
        #print(dp)
        return dp[-1]