Leetcode simple question ring and rod

subject

A total of n Annulus , The color of the ring can be red 、 green 、 One of the blue . These rings are distributed across 10 The root number is 0 To 9 On the rod .
Give you a length of 2n String rings , It means n The distribution of rings on the rod .rings Every two characters in the form of a The color position is right , Used to describe each ring ：
The first i Right first The character represents the second i A ring Color （‘R’、‘G’、‘B’）.
The first i Right the second The character represents the second i A ring Location , That is, on which rod （‘0’ To ‘9’）.
for example ,“R3G2B1” Express ： share n == 3 Annulus , The red ring is numbered 3 On the rod , The green ring is numbered 2 On the rod , The blue ring is numbered 1 On the rod .
Find all sets All three colors The rod of the ring , And return the number of such rods .
Example 1：
Input ：rings = “B0B6G0R6R0R6G9”
Output ：1
explain ：

• Number 0 On the pole 3 Annulus , Gather all colors ： red 、 green 、 blue .
• Number 6 On the pole 3 Annulus , But only red 、 Blue two colors .
• Number 9 There is only 1 A green ring .
  therefore , The number of rods with all three color rings is 1 .
  Example 2：
  Input ：rings = “B0R0G0R9R0B0G0”
  Output ：1
  explain ：
• Number 0 On the pole 6 Annulus , Gather all colors ： red 、 green 、 blue .
• Number 9 There is only 1 A red ring .
  therefore , The number of rods with all three color rings is 1 .
  Example 3：
  Input ：rings = “G4”
  Output ：0
  explain ：
  Only one ring was given , therefore , There is no rod that gathers all three color rings .
  Tips ：
  rings.length == 2 * n
  1 <= n <= 100
  Such as i yes even numbers , be rings[i] The value of can be taken as ‘R’、‘G’ or ‘B’（ Subscript from 0 Start counting ）
  Such as i yes Odd number , be rings[i] The value of can be taken as ‘0’ To ‘9’ A number in （ Subscript from 0 Start counting ）
  source ： Power button （LeetCode）

Their thinking

   This problem only needs to count the colors on each pole , You can solve this problem by embedding a hash table in a dictionary .

class Solution:
    def countPoints(self, rings: str) -> int:
        d={
    '0':[],'1':[],'2':[],'3':[],'4':[],'5':[],'6':[],'7':[],'8':[],'9':[]}
        for i in range(len(rings)):
            if rings[i].isdigit():
                if rings[i-1] not in d[rings[i]]:
                    d[rings[i]].append(rings[i-1])
        count=0
        for i in d.values():
            if len(i)==3:
                count+=1
        return count