1.重复的子字符串classSolution{public:voidgetNext(vectorintnext,conststring s){intj0;next[j]0;for(inti1;is.size();i){while(j-10s[i]!s[j]){jnext[j-1];}if(s[i]s[j]){j;next[i]j;}else{next[i]0;}}}boolrepeatedSubstringPattern(string s){vectorintnext(s.size(),0);getNext(next,s);intmaxindenextnext[s.size()-1];returnmaxindenext!0s.size()%(s.size()-maxindenext)0;}};2.找出字符串中第一个匹配项的下标思路先确定needlenext数组 然后遍历next 如果当前元素和needle【j】 相等 继续遍历 不等循环回退jnext[j-1];再次判断是否相等 不相等继续回退到j0相等的话直接j;while(j-10haystack[i]!needle[j]){jnext[j-1];}if(haystack[i]needle[j]){j;}完整代码classSolution{voidgetNext(vectorintnext,string needle){intj0;next[j]0;for(inti1;ineedle.size();i){while(j-10needle[i]!needle[j]){jnext[j-1];}if(needle[i]needle[j]){j;}next[i]j;}}public:intstrStr(string haystack,string needle){vectorintnext(needle.size(),0);getNext(next,needle);intj0;for(inti0;ihaystack.size();i){while(j-10haystack[i]!needle[j]){jnext[j-1];}if(haystack[i]needle[j]){j;}if(jneedle.size()){returni-needle.size()1;}}return-1;}};