[leetcode] 1642. Furthest Building You Can Reach

Description

You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.

You start your journey from building 0 and move to the next building by possibly using bricks or ladders.

While moving from building i to building i+1 (0-indexed),

If the current building’s height is greater than or equal to the next building’s height, you do not need a ladder or bricks.
If the current building’s height is less than the next building’s height, you can either use one ladder or (h[i+1] - h[i]) bricks.
Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

Example 1:

Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
Output: 4
Explanation: Starting at building 0, you can follow these steps:
- Go to building 1 without using ladders nor bricks since 4 >= 2.
- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
- Go to building 3 without using ladders nor bricks since 7 >= 6.
- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
It is impossible to go beyond building 4 because you do not have any more bricks or ladders.

Example 2:

Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
Output: 7

Example 3:

Input: heights = [14,3,19,3], bricks = 17, ladders = 0
Output: 3

Constraints:

• 1 <= heights.length <= 105
• 1 <= heights[i] <= 106
• 0 <= bricks <= 109
• 0 <= ladders <= heights.length

分析

題目的意思是：給你一個陣列，表示樓層的高，你又bricks塊磚和ladders個梯子，問能夠到達的最遠距離。這道題如果你知道用優先佇列的話，就好做了。剩下的就是根據

程式碼

class Solution:
    def furthestBuilding(self, heights: List[int], bricks: int, ladders: int) -> int:
        cnt=0
        heap=[]
        for i in range(1,len(heights)):
            if(heights[i]>heights[i-1]):
                cnt=heights[i]-heights[i-1]
                heapq.heappush(heap,cnt)
                if(len(heap)>ladders):
                    bricks-=heapq.heappop(heap)
                if(bricks<0):
                    return i-1
        return len(heights)-1

參考文獻

[LeetCode] [Java/C++/Python] Priority Queue