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81 lines
2.7 KiB
Markdown
81 lines
2.7 KiB
Markdown
# [1742. Maximum Number of Balls in a Box](https://leetcode.com/problems/maximum-number-of-balls-in-a-box/)
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## 题目
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You are working in a ball factory where you have `n` balls numbered from `lowLimit` up to `highLimit` **inclusive** (i.e., `n == highLimit - lowLimit + 1`), and an infinite number of boxes numbered from `1` to `infinity`.
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Your job at this factory is to put each ball in the box with a number equal to the sum of digits of the ball's number. For example, the ball number `321` will be put in the box number `3 + 2 + 1 = 6` and the ball number `10` will be put in the box number `1 + 0 = 1`.
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Given two integers `lowLimit` and `highLimit`, return *the number of balls in the box with the most balls.*
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**Example 1:**
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```
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Input: lowLimit = 1, highLimit = 10
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Output: 2
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Explanation:
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Box Number: 1 2 3 4 5 6 7 8 9 10 11 ...
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Ball Count: 2 1 1 1 1 1 1 1 1 0 0 ...
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Box 1 has the most number of balls with 2 balls.
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```
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**Example 2:**
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```
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Input: lowLimit = 5, highLimit = 15
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Output: 2
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Explanation:
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Box Number: 1 2 3 4 5 6 7 8 9 10 11 ...
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Ball Count: 1 1 1 1 2 2 1 1 1 0 0 ...
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Boxes 5 and 6 have the most number of balls with 2 balls in each.
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```
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**Example 3:**
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```
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Input: lowLimit = 19, highLimit = 28
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Output: 2
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Explanation:
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Box Number: 1 2 3 4 5 6 7 8 9 10 11 12 ...
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Ball Count: 0 1 1 1 1 1 1 1 1 2 0 0 ...
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Box 10 has the most number of balls with 2 balls.
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```
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**Constraints:**
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- `1 <= lowLimit <= highLimit <= 10^5`
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## 题目大意
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你在一家生产小球的玩具厂工作，有 n 个小球，编号从 lowLimit 开始，到 highLimit 结束（包括 lowLimit 和 highLimit ，即 n == highLimit - lowLimit + 1）。另有无限数量的盒子，编号从 1 到 infinity 。你的工作是将每个小球放入盒子中，其中盒子的编号应当等于小球编号上每位数字的和。例如，编号 321 的小球应当放入编号 3 + 2 + 1 = 6 的盒子，而编号 10 的小球应当放入编号 1 + 0 = 1 的盒子。
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给你两个整数 lowLimit 和 highLimit ，返回放有最多小球的盒子中的小球数量。如果有多个盒子都满足放有最多小球，只需返回其中任一盒子的小球数量。
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## 解题思路
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- 简单题。循环遍历一遍数组，依次计算出所有小球的编号各位数字累加和，并且动态维护放有小球最多的数目。循环结束，输出最多小球个数即可。
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## 代码
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```go
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package leetcode
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func countBalls(lowLimit int, highLimit int) int {
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buckets, maxBall := [46]int{}, 0
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for i := lowLimit; i <= highLimit; i++ {
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t := 0
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for j := i; j > 0; {
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t += j % 10
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j = j / 10
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}
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buckets[t]++
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if buckets[t] > maxBall {
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maxBall = buckets[t]
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}
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}
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return maxBall
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}
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``` |