Furthest Building You Can Reach

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.

思路：greedy思路，就是用maxheap记录当前用掉的bricks，因为ladder可以填充任意的高度，所以很珍贵，首先用bricks填充，然后如果不够了，那么就用ladder去填充之前用过的最大的bricks。

class Solution {
    public int furthestBuilding(int[] heights, int bricks, int ladders) {
        PriorityQueue<Integer> maxheap = new PriorityQueue<Integer>((a, b) -> (b - a));
        for(int i = 0; i < heights.length; i++) {
            if(i + 1 < heights.length && heights[i] >= heights[i + 1]) {
                continue;
            } 
            if(i + 1 < heights.length && heights[i] < heights[i + 1]) {
                int diff = heights[i + 1] - heights[i];
                // if bricks is enough, first use bricks.
                if(bricks >= diff) {
                    bricks -= diff;
                    maxheap.offer(diff);
                } else {
                    // not enough bricks, try to use ladder fill largest bricks gap;
                    if(ladders > 0) {
                        if(!maxheap.isEmpty()) {
                            bricks += maxheap.poll();
                            // if bricks is enough, first use bricks. 
                            if(bricks >= diff) {
                               bricks -= diff;
                               maxheap.offer(diff); 
                            }
                        }
                        ladders--;
                    } else {
                        return i;
                    }
                }
            }
        }
        return heights.length - 1;
    }
}