Two houses with different colors and the farthest distance

subject

There are... In the street n The houses are neatly lined up , Every house is painted in beautiful colors . I'll give you a subscript from 0 Start with a length of n Array of integers for colors , among colors[i] It means the first one i The color of the house .
return Two buildings Color Different Between houses Maximum distance .
The first i This house and the second j The distance between the houses is abs(i - j) , among abs(x) yes x The absolute value of .
Example 1：

Input ：colors = [1,1,1,6,1,1,1]
Output ：3
explain ： Above picture , Color 1 The logo is blue , Color 6 The logo is red .
The two houses with different colors and the farthest distance are houses 0 And the house 3 .
house 0 The color of is color 1 , house 3 The color of is color 6 . The distance between the two houses is abs(0 - 3) = 3 .
Be careful , house 3 And the house 6 It can also produce the best answer .
Example 2：

Input ：colors = [1,8,3,8,3]
Output ：4
explain ： Above picture , Color 1 The logo is blue , Color 8 The logo is yellow , Color 3 The logo is green .
The two houses with different colors and the farthest distance are houses 0 And the house 4 .
house 0 The color of is color 1 , house 4 The color of is color 3 . The distance between the two houses is abs(0 - 4) = 4 .
Example 3：
Input ：colors = [0,1]
Output ：1
explain ： The two houses with different colors and the farthest distance are houses 0 And the house 1 .
house 0 The color of is color 0 , house 1 The color of is color 1 . The distance between the two houses is abs(0 - 1) = 1 .
Tips ：
n == colors.length
2 <= n <= 100
0 <= colors[i] <= 100
The generated test data meets At least There is 2 A house of different colors
source ： Power button （LeetCode）

Their thinking

   First, find out how many different colors there are , Then traverse from the front to find the position of the color , Then we traverse the tail to find the first position of a color different from this color , This should be the color 1 To color 2 The maximum distance .

class Solution:
    def maxDistance(self, colors: List[int]) -> int:
        color=set(colors)
        MAX=0
        for i in color:
            for j in range(len(colors)):
                if i==colors[j]:
                    break
            for k in range(len(colors)-1,-1,-1):
                if colors[k]!=i:
                    break
            if k-j>MAX:
                MAX=k-j
        return MAX