代码随想录算法训练营第9天| ●28. 实现 strStr() ●459.重复的子字符串 ●字符串总结 ●双指针回顾 -526互联

28.找出字符串中第一个匹配项的下标

mydemo--(my thought)--(falied)

class Solution {
public:
    int strStr(string haystack, string needle) {
        
        for(int i=0; i<haystack.size(); i++)
        {
            if(haystack[i] != needle[0])    continue;
            int j=0;
            int start = i;
            while(haystack[i]==needle[j])
            {
                if(j == needle.size()-1) return start;
                j++;
                i++;
            }
        }

        return -1;
    }
};

mydemoV2--(暴力解法，自己思路)--(successed)

class Solution {
public:
    int strStr(string haystack, string needle) {
        
        int start = 0;
        int len1 = haystack.size();
        int len2 = needle.size();

        while(start < len1)
        {
            int i = start;
            int j = 0;
            while(haystack[i] == needle[j] && i < len1 && j < len2)
            {
                if(j == len2 - 1)    return start;
                i++;
                j++;
            }
            start++;
        }

        return -1;
    }
};

时间复杂度 O(n*m)

前缀：包含首字母，不包含尾字母的所有字符串

后缀：包含尾字母，不包含首字母的所有字符串

卡哥demo--KMP算法

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;
        int next[needle.size()];
        getNext(next, 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;
    }
};

459.重复的子字符串

卡哥代码

KMP思路

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;
    }
};