python leetcode 821. Shortest Distance to a Character

最新推荐文章于 2019-11-21 16:47:17 发布

DreamerLHQ

最新推荐文章于 2019-11-21 16:47:17 发布

阅读量800

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode_python 文章标签： leetcode python

本文链接：https://blog.youkuaiyun.com/DreamerLHQ/article/details/80116531

leetcode_python 专栏收录该内容

71 篇文章

订阅专栏

本文介绍了一种算法，用于在给定字符串中找到特定字符并计算每个位置到该字符的最短距离。通过实例演示了如何实现这一功能。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given a string S and a character C, return an array of integers representing the shortest distance from the character C in the string.

找出字符串中与给定字符的最短距离。

Example 1:

Input: S = "loveleetcode", C = 'e'
Output: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]

Note:

S string length is in [1, 10000].
C is a single character, and guaranteed to be in string S.
All letters in S and C are lowercase.

class Solution:
def shortestToChar(self, S, C):
"""
:type S: str
:type C: str
:rtype: List[int]
"""
index_nums = []
for i, letter in enumerate(S):
if letter == C:
index_nums.append(i)

ans = []
for j, word in enumerate(S):
distance = [abs(j - index_num) for index_num in index_nums]
ans.append(min(distance))

return ans