描述
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.
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.
Given two integers lowLimit and highLimit, return the number of balls in the box with the most balls.
Example 1:
Input: lowLimit = 1, highLimit = 10
Output: 2
Explanation:
Box Number: 1 2 3 4 5 6 7 8 9 10 11 ...
Ball Count: 2 1 1 1 1 1 1 1 1 0 0 ...
Box 1 has the most number of balls with 2 balls.
Example 2:
Input: lowLimit = 5, highLimit = 15
Output: 2
Explanation:
Box Number: 1 2 3 4 5 6 7 8 9 10 11 ...
Ball Count: 1 1 1 1 2 2 1 1 1 0 0 ...
Boxes 5 and 6 have the most number of balls with 2 balls in each.
Example 3:
Input: lowLimit = 19, highLimit = 28
Output: 2
Explanation:
Box Number: 1 2 3 4 5 6 7 8 9 10 11 12 ...
Ball Count: 0 1 1 1 1 1 1 1 1 2 0 0 ...
Box 10 has the most number of balls with 2 balls.
Note:
1 <= lowLimit <= highLimit <= 10^5
解析
根据题意，只需要遍历从 lowLimit 到 highLimit 的数值 item ，然后使用字典 balls 保存其放的球数，以 item 每个位数上的和为 key ，以可以存放的球数为 value ，遍历结束即可得到每个 key 的结果，然后找出最大值即可。
解答
class Solution(object):
def countBalls(self, lowLimit, highLimit):
"""
:type lowLimit: int
:type highLimit: int
:rtype: int
"""
balls = {}
def sum(num):
result = 0
while num>0:
result += num%10
num=num//10
return result
for item in range(lowLimit,highLimit+1):
tmp = sum(item)
if tmp not in balls:
balls[tmp]=1
else:
balls[tmp]+=1
return max(balls.values())
运行结果
Runtime: 504 ms, faster than 79.76% of Python online submissions for Maximum Number of Balls in a Box.
Memory Usage: 15.9 MB, less than 86.84% of Python online submissions for Maximum Number of Balls in a Box.
原题链接：leetcode.com/problems/ma…
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