我报名参加金石计划1期挑战——瓜分10万奖池，这是我的第10篇文章，点击查看活动详情
描述
You are given a 0-indexed string s consisting of only lowercase English letters, where each letter in s appears exactly twice. You are also given a 0-indexed integer array distance of length 26. Each letter in the alphabet is numbered from 0 to 25 (i.e. 'a' -> 0, 'b' -> 1, 'c' -> 2, ... , 'z' -> 25). In a well-spaced string, the number of letters between the two occurrences of the ith letter is distance[i]. If the ith letter does not appear in s, then distance[i] can be ignored.
Return true if s is a well-spaced string, otherwise return false.
Example 1:
Input: s = "abaccb", distance = [1,3,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: true
Explanation:
- 'a' appears at indices 0 and 2 so it satisfies distance[0] = 1.
- 'b' appears at indices 1 and 5 so it satisfies distance[1] = 3.
- 'c' appears at indices 3 and 4 so it satisfies distance[2] = 0.
Note that distance[3] = 5, but since 'd' does not appear in s, it can be ignored.
Return true because s is a well-spaced string.
Example 2:
Input: s = "aa", distance = [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
Output: false
Explanation:
- 'a' appears at indices 0 and 1 so there are zero letters between them.
Because distance[0] = 1, s is not a well-spaced string.
Note:
2 <= s.length <= 52
s consists only of lowercase English letters.
Each letter appears in s exactly twice.
distance.length == 26
0 <= distance[i] <= 50
解析
根据题意，给定一个仅由小写英文字母组成的索引为 0 的字符串 s ，其中 s 中的每个字母恰好出现两次。 还给定一个长度为 26 的 0 索引整数数组 distance 。字母表中的每个字母编号从 0 到 25（即 'a' -> 0、'b' -> 1、'c' -> 2、. .. , 'z' -> 25)。在间隔良好的字符串中，第 ith 个字母的两次出现之间的字母个数是 distance[i]。 如果第 ith 字母没有出现在 s 中，那么 distance[i] 可以忽略。如果 s 是一个间隔良好的字符串，则返回 true，否则返回 false。
其实这道题就是考察一个很简单的数组遍历，因为题目中已经给出了字符肯定会有两次，所以我们直接使用一个字典 d 将出现的字符及其位置都存好，然后只要从 a 开始遍历到 z ，只要该字符出现在 d 中（不存在的就不用管了），我们判断如果该字符在 d 中存的距离不等于 distance 存的距离，那么就直接返回 False ，否则遍历结束直接返回 True 。
时间复杂度为 O(N) ，空间复杂度为 O(N) 。
解答
class Solution(object):
def checkDistances(self, s, distance):
"""
:type s: str
:type distance: List[int]
:rtype: bool
"""
d = collections.defaultdict(list)
for i,c in enumerate(s):
d[c].append(i)
for i in range(26):
c = chr(i+97)
if c in d:
if distance[i] != d[c][1]- d[c][0] - 1:
return False
return True
运行结果
335 / 335 test cases passed.
Status: Accepted
Runtime: 47 ms
Memory Usage: 13.6 MB
原题链接
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