★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
?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: 8Explanation: 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(@H_301_98@"@H_301_98@#@H_301_98@") 5 var k:Int = 0 6 var c:Character = @H_301_98@"@H_301_98@#@H_301_98@" 7 var ans:Int = 0 8 for i in 0..<arr.count 9 {10 if arr[i] == c11 {12 k += 113 }14 else15 {16 if c != @H_301_98@"@H_301_98@#@H_301_98@"17 {18 ans += k * (k + 1) / 219 }20 c = arr[i]21 k = 122 }23 }24 return ans25 }26 }总结
以上是内存溢出为你收集整理的[Swift]LeetCode1180. 统计只含单一字母的子串 | Count Substrings with Only One Distinct Letter全部内容,希望文章能够帮你解决[Swift]LeetCode1180. 统计只含单一字母的子串 | Count Substrings with Only One Distinct Letter所遇到的程序开发问题。
如果觉得内存溢出网站内容还不错,欢迎将内存溢出网站推荐给程序员好友。
欢迎分享,转载请注明来源:内存溢出
评论列表(0条)