代码随想录算法训练营第37期 第九天 | LeetCode28. 实现 strStr()、459.重复的子字符串

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一、28. 实现 strStr()

解题代码C++：

class Solution {
public:void getNext(int* next, const string& s) {int j = 0;next[0] = 0;for(int i = 1; i < s.size(); i++) {while (j > 0 && s[i] != s[j]) {j = next[j - 1];}if (s[i] == s[j]) {j++;}next[i] = j;}}int strStr(string haystack, string needle) {if (needle.size() == 0) {return 0;}vector<int> next(needle.size());getNext(&next[0], needle);int j = 0;for (int i = 0; i < haystack.size(); i++) {while(j > 0 && haystack[i] != needle[j]) {j = next[j - 1];}if (haystack[i] == needle[j]) {j++;}if (j == needle.size() ) {return (i - needle.size() + 1);}}return -1;}
};

题目链接/文章讲解/视频讲解：
https://programmercarl.com/0028.%E5%AE%9E%E7%8E%B0strStr.html

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二、459.重复的子字符串

解题代码C++：

class Solution {
public:void getNext (int* next, const string& s){next[0] = 0;int j = 0;for(int i = 1;i < s.size(); i++){while(j > 0 && s[i] != s[j]) {j = next[j - 1];}if(s[i] == s[j]) {j++;}next[i] = j;}}bool repeatedSubstringPattern (string s) {if (s.size() == 0) {return false;}int next[s.size()];getNext(next, s);int len = s.size();if (next[len - 1] != 0 && len % (len - (next[len - 1] )) == 0) {return true;}return false;}
};

题目链接/文章讲解/视频讲解：
https://programmercarl.com/0459.%E9%87%8D%E5%A4%8D%E7%9A%84%E5%AD%90%E5%AD%97%E7%AC%A6%E4%B8%B2.html