LEETCODE 1564. Put Boxes Into the Warehouse I 解题思路分析

题目大意:

将箱子放进仓库1

Given two arrays of positive integers boxes and warehouse representing the heights of some boxes of unit width, and the heights of n rooms in a warehouse, respectively. The warehouse’s rooms are labeled from 0 to n - 1 from left to right where warehouse[i] (0-indexed) is the height of the ith room.

Boxes are put into the warehouse by the following rules:

  • Boxes can’t be piled up.
  • You can rearrange the order of the boxes.
  • Boxes can only be pushed into the warehouse from left to right only.
  • If the height of some room in the warehouse is less than the height of a box, then the box will be stopped before that room, so are the boxes behind it.

Return the maximum number of boxes you can put into the warehouse.

示例1:

输入: boxes = [4,3,4,1], warehouse = [5,3,3,4,1]
输出: 3
解释: 

We can first put the box of height 1 in room 4. Then we can put the box of height 3 in either of the 3 rooms 1, 2, or 3. Lastly, we can put one box of height 4 in room 0. There is no way we can fit all 4 boxes in the warehouse.

示例2:

输入: boxes = [1,2,2,3,4], warehouse = [3,4,1,2]
输出: 3
解释: 

Notice that it's not possible to put the box of height 4 into the warehouse since it cannot pass the first room of height 3. Also, for the last two rooms, 2 and 3, only boxes of height 1 can fit. We can fit 3 boxes maximum as shown above. The yellow box can also be put in room 2 instead. Swapping the orange and green boxes is also valid, or swapping one of them with the red box.

Example 3:

Input: boxes = [1,2,3], warehouse = [1,2,3,4]
Output: 1
Explanation: Since the first room in the warehouse is of height 1, we can only put boxes of height 1.

Example 4:

Input: boxes = [4,5,6], warehouse = [3,3,3,3,3]
Output: 0

Constraints:

  • n == warehouse.length
  • 1 <= boxes.length, warehouse.length <= 10^5
  • 1 <= boxes[i], warehouse[i] <= 10^9

如果想查看本题目是哪家公司的面试题,请参考以下免费链接: https://leetcode.jp/problemdetail.php?id=1564

解题思路分析:

这道题是向仓库内放箱子,题目给出的仓库很奇怪,它的屋顶高度是错落无序的。由于我们只能从仓库左边向内放入,因此,如果想放入更多的箱子,应该使得先进入的箱子应该尽量放到仓库的靠右的位置。而一个箱子能否放到某个位置i上,取决于仓库的最左边(下标0)到当前位置下标i的区间内,是否高度都大于等于当前箱子的高度,如果是,才能保证箱子能顺利从仓库的左边入口移动到下标i处。解题时:

  1. 我们需要先对数据进行预处理,即算出仓库的每一个位置上到入口间的最小高度,即leftMinHeight[i]表示下标0到i间的最小高度。计算这个高度很简单,我们使用一个循环,从仓库下标0循环至末尾。下标0处,leftMinHeight[0]等于仓库位置0处的高度warehouse[0],而其他位置leftMinHeight[i]则等于: min(warehouse[i],leftMinHeight[i-1])
  2. 接下来我们对箱子按照高度进行排序
  3. 将当前箱子尽量向仓库的右边放。即使用双指针,初始时,箱子指针指向下标0(高度最小的箱子),仓库指针指向仓库最右端。如果当前箱子的高度小于等于leftMinHeight[i],即,当前箱子可以从入口移动到仓库当前位置,此时,返回结果加一,同时箱子指针向右移动一位,仓库指针向左移动一位。反之,代表当前箱子无法顺利移动到仓库当前位置,那么,我们只能将仓库位置向左移动一位。重复上述操作,直到仓库位置小于0,或者箱子下标大于等于箱子个数时为止。

实现代码(Java版本):

public int maxBoxesInWarehouse(int[] boxes, int[] warehouse) {
    int[] leftMinHeight = new int[warehouse.length]; // 统计仓库当前位置到入口间的最小高度
    for(int i=0;i<warehouse.length;i++){ // 循环仓库每个位置
        if(i==0) leftMinHeight[i]=warehouse[i]; // 如果下标为0,最小高度为当前高度
        // 否则最小高度为当前高度与前移最小高度的最小值
        else leftMinHeight[i]=Math.min(leftMinHeight[i-1],warehouse[i]);
    }
    Arrays.sort(boxes); // 箱子按照高度排序
    int res=0; // 返回结果
    int boxIndex=0; // 箱子下标
    int houseIndex=leftMinHeight.length-1; // 仓库下标
    while(boxIndex<boxes.length&&houseIndex>=0){
        // 如果箱子可以移动到当前位置
        if(boxes[boxIndex]<=leftMinHeight[houseIndex]){
            res++; // 返回结果加一
            boxIndex++; // 箱子下标加一(继续放下一个箱子)
            houseIndex--; // 仓库下标减一(继续从当前位置左边开始尝试放箱子)
        }else{ // 无法移动到当前位置
            houseIndex--; // 仓库下标减一(继续从当前位置左边开始尝试放箱子)
        }
    }
    return res;
}

本题解法执行时间为16ms。

Runtime: 16 ms, faster than 100.00% of Java online submissions for Put Boxes Into the Warehouse I.

Memory Usage: 52.7 MB, less than 100.00% of Java online submissions for Put Boxes Into the Warehouse I.

实现代码(Kotlin版本):

fun maxBoxesInWarehouse(boxes: IntArray, warehouse: IntArray): Int {
    var leftMinHeight = mutableListOf<Int>()
    for(i in 0..(warehouse.size-1)){
        leftMinHeight.add(
            if(i==0)
                warehouse[0]
              else
                Math.min(warehouse[i],leftMinHeight[i-1]))
    }
    Arrays.sort(boxes);
    var res = 0
    var indexBox = 0
    var indexHouse = warehouse.size-1
    while(indexHouse>=0&&indexBox<leftMinHeight.size){
        if(boxes[indexBox]<=leftMinHeight[indexHouse]){
            res++
            indexBox++
            indexHouse--
        }else{
            indexHouse--
        }
    }
    return res;
}
本网站文章均为原创内容,并可随意转载,但请标明本文链接
如有任何疑问可在文章底部留言。为了防止恶意评论,本博客现已开启留言审核功能。但是博主会在后台第一时间看到您的留言,并会在第一时间对您的留言进行回复!欢迎交流!
本文链接: https://leetcode.jp/leetcode-1564-put-boxes-into-the-warehouse-i-解题思路分析/
此条目发表在leetcode分类目录,贴了, , , , , 标签。将固定链接加入收藏夹。

发表评论

电子邮件地址不会被公开。