★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★ ?微信公众号:为敢(WeiGanTechnologies) ?博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
?微信公众号:为敢(WeiGanTechnologies)
?博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)
?GitHub地址:https://github.com/strengthen/LeetCode
?原文地址:https://www.cnblogs.com/strengthen/p/11484244.html
?如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
?原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
Given a string S
, return the number of substrings that have only one distinct letter.
Example 1:
Input: S = "aaaba" Output: 8 Explanation: The substrings with one distinct letter are "aaa", "aa", "a", "b". "aaa" occurs 1 time. "aa" occurs 2 times. "a" occurs 4 times. "b" occurs 1 time. So the answer is 1 + 2 + 4 + 1 = 8.
Example 2:
Input: S = "aaaaaaaaaa" Output: 55
Constraints:
1 <= S.length <= 1000
S[i]
consists of only lowercase English letters.
给你一个字符串 S
,返回只含 单一字母 的子串个数。
示例 1:
输入: "aaaba" 输出: 8 解释: 只含单一字母的子串分别是 "aaa", "aa", "a", "b"。 "aaa" 出现 1 次。 "aa" 出现 2 次。 "a" 出现 4 次。 "b" 出现 1 次。 所以答案是 1 + 2 + 4 + 1 = 8。
示例 2:
输入: "aaaaaaaaaa" 输出: 55
提示:
1 <= S.length <= 1000
S[i]
仅由小写英文字母组成。
Runtime: 4 ms Memory Usage: 21 MB
1 class Solution { 2 func countLetters(_ S: String) -> Int { 3 var arr:[Character] = Array(S) 4 arr.append("#") 5 var k:Int = 0 6 var c:Character = "#" 7 var ans:Int = 0 8 for i in 0..<arr.count 9 { 10 if arr[i] == c 11 { 12 k += 1 13 } 14 else 15 { 16 if c != "#" 17 { 18 ans += k * (k + 1) / 2 19 } 20 c = arr[i] 21 k = 1 22 } 23 } 24 return ans 25 } 26 }