文章作者:Tyan 博客:noahsnail.com | CSDN | 简书
1. Description
2. Solution
**解析:**Version 1,两层循环遍历,O(N^2)。
- Version 1
class Solution:
def maxDistance(self, colors: List[int]) -> int:
distance = 0
length = len(colors)
for i in range(length):
for j in range(i 1, length):
if colors[i] != colors[j]:
distance = max(distance, j - i)
return distance
**解析:**Version 2,贪心算法,从右往左找与左边第一个不同颜色的房子,从左往右找与右边第一个不同颜色的房子,取距离最大的一个,O(N)。
- Version 2
class Solution:
def maxDistance(self, colors: List[int]) -> int:
distance = 0
length = len(colors)
for i in range(length-1, -1, -1):
if colors[i] != colors[0]:
distance = max(distance, i)
break
for i in range(length):
if colors[i] != colors[length - 1]:
distance = max(distance, length - 1 - i)
break
return distance
**解析:**Version 3,Version 2的另一个版本,通过一次循环完成,O(N)。
- Version 3
class Solution:
def maxDistance(self, colors: List[int]) -> int:
distance = 0
length = len(colors)
for index, color in enumerate(colors):
if color != colors[0]:
distance = max(distance, index)
if color != colors[length-1]:
distance = max(distance, length - 1 - index)
return distance
Reference
- https://leetcode.com/problems/two-furthest-houses-with-different-colors/