heap系列

1642. Furthest Building You Can Reach Medium

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

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        //1.define a heap
        PriorityQueue<Integer> pq = new PriorityQueue<>((x,y)->x-y);
        //2.for loop the heights,
        //2.1 if heap size is bigger than ladders, than we need to poll the least height one to use bricks
        for(int i=0;i<heights.length;i++){
            if(i>0 && heights[i]>heights[i-1]){
                int diff = heights[i]-heights[i-1];
                pq.offer(diff);
                if(pq.size()>ladders){
                    bricks-=pq.poll();
                }
                if(bricks<0) return i-1;
            }
        }
        return heights.length-1;
    }
}

 

标签：building,系列,bricks,heights,ladders,int,heap,using
来源： https://www.cnblogs.com/cynrjy/p/15762006.html