力扣题-12.7

[力扣刷题攻略] Re：从零开始的力扣刷题生活

力扣题1：467. 环绕字符串中唯一的子字符串

解题思想：记录下以字母s[i]结尾的最大的字串个数，然后统计a-z每个字母结尾的最大字串的个数进行i相加

class Solution(object):def findSubstringInWraproundString(self, s):""":type s: str:rtype: int"""dp = defaultdict(int)k = 0for i in range(len(s)):if i>0 and (ord(s[i])-ord(s[i-1])) %26 == 1:k += 1else:k = 1dp[s[i]] = max(dp[s[i]],k)return sum(dp.values())

class Solution {
public:int findSubstringInWraproundString(string s) {unordered_map<char, int> dp;int k=0;for(int i=0;i<s.size();i++){if (i > 0 && (s[i] - s[i - 1] + 26) % 26 == 1) {k++;}else{k = 1;}dp[s[i]] = max(dp[s[i]], k);}int result = 0;for (const auto& entry : dp) {result += entry.second;}return result;}
};