Re: [閒聊] 每日LeetCode

看板Marginalman作者ZooseWu (動物園公告)時間2年前 (2023/11/09 14:28)推噓2(2推 0噓 1→)

留言3則, 3人參與討論串500/719 (看更多)

※ 引述《Neuenmuller (蘇菲・諾伊恩謬拉)》之銘言： : 1759. Count Number of Homogenous Substrings : 找相同字元的substring 數各種substring總共出現多少次 : ex: input = "aabbbca" : "a": 3 : "aa": 1 : "b": 3 : "bb": 2 : "bbb": 1 : "c": 1 : 總共 11 思路差不多 Intuition 每個字母都與其他字母無關因此將整個字串根據相連的字母分成子字串計算 Approach 假設現在有一個子字串aaa 我們去計算它擁有的子字串: index = 0: "a" index = 1: "a", "aa" index = 2: "a", "aa", "aaa" 從規律可以看到第n次出現 = 新的子字串總數所以每當長度+1我們就把答案加上長度就好 TS Code: const mod = 1000000007 function countHomogenous (s: string): number { let lastChar = "" let subStringLength = 1 let answer = 0 for (let i = 0; i < s.length; i++) { if (s[i] === lastChar) { subStringLength++ } else { lastChar = s[i] subStringLength = 1 } answer = (answer + subStringLength) % mod } return answer } C# Code: public class Solution { public int CountHomogenous(string s) { int mod = 1000000007; char lastChar = '\n'; var subStringLength = 0; var answer = 0; for (int i = 0; i < s.Length; i++) { if (s[i] == lastChar) { subStringLength++; } else { lastChar = s[i]; subStringLength = 1; } answer = (answer + subStringLength) % mod; } return answer; } } Go Code: func countHomogenous(s string) int { mod := 1000000007 lastChar := s[0] subStringLength := 0 answer := 0 for i := 0; i < len(s); i++ { if s[i] == lastChar { subStringLength++ } else { lastChar = s[i] subStringLength = 1 } answer = (answer + subStringLength) % mod } return answer } -- Zoosewu Yoututbe顯示PTT推文可以在各個網站追實況或Live時使用預覽圖: https://i.imgur.com/ZhtXdAJ.png