题目

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:

q4.gif

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

Example 2:

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

Example 3:

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

Constraints:

  • 1 <= heights.length <= 10^5
  • 1 <= heights[i] <= 10^6
  • 0 <= bricks <= 10^9
  • 0 <= ladders <= heights.length

题意

给定一个高度数组，从左到右移动。如果当前高度大于等于下一个高度，可以直接移动；如果当前高度小于下一个高度，记差为diff，则需要diff块砖或一个梯子才能移动到下一个位置，求能移动到的最远的位置。

思路

贪心。因为一个梯子能代替任意块砖，应该把梯子用在移动过程中差值最大的地方。做法如下：计算差值diff，如果小于等于0，直接移动；如果大于0，优先使用brick，并将diff加入优先队列，如果brick不够，则消耗一个ladder，置换出已消耗的最大diff返还给brick。

代码实现

Java

  1. class Solution {
  2.     public int furthestBuilding(int[] heights, int bricks, int ladders) {
  3.         int index = 0;
  4.         Queue<Integer> top = new PriorityQueue<>((a, b) -> b - a);
  6.         while (index + 1 < heights.length) {
  7.             int diff = heights[index + 1] - heights[index];
  9.             if (diff <= 0) {
  10.                 index++;
  11.             } else if (bricks >= diff) {
  12.                 bricks -= diff;
  13.                 top.offer(diff);
  14.                 index++;
  15.             } else if (ladders == 0) {
  16.                 break;
  17.             } else if (top.isEmpty() || top.peek() <= diff) {
  18.                 ladders--;
  19.                 index++;
  20.             } else {
  21.                 ladders--;
  22.                 bricks += top.poll();
  23.                 if (bricks >= diff) {
  24.                     bricks -= diff;
  25.                     top.offer(diff);
  26.                     index++;
  27.                 }
  28.             }
  29.         }
  31.         return index;
  32.     }
  33. }