kmp算法（完结）-优快云博客

1.重复的子字符串

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

    }
    bool repeatedSubstringPattern(string s) {
        vector<int> next(s.size(),0);
        getNext(next,s);
        int maxindenext = next[s.size()-1];
        return maxindenext!=0&&s.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j=next[j-1];再次判断是否相等不相等继续回退到j=0相等的话直接j++;

            while(j-1>=0&&haystack[i]!=needle[j]){
                j = next[j-1];
            }
            if(haystack[i] == needle[j]){
                j++;
            }

完整代码

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

public:
    int strStr(string haystack, string needle) {
        vector<int > next(needle.size(),0);
        getNext(next,needle);
        int j =0;
        for(int i=0;i<haystack.size();i++){
            while(j-1>=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;
        
    }
};