28.找出字符串中第一个匹配项的下标【力扣】KMP前缀表 ≈ find() 函数、暴力解法

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class Solution {
public:
//得到前缀表void getNext(int *next,string needle){int j=0;for(int i=1;i<needle.size();i++){while(j>0 && needle[j]!=needle[i]) j=next[j-1];//**j>0**=>j==0是出口if(needle[i]==needle[j]) j++;next[i]==j;//若写入if中，则该行意思变为当匹配时才给next赋值，没能顾及到回退时的赋值}}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];//运用前缀表if(haystack[i]==needle[j]) j++;if(j==needle.size()) return i-needle.size()+1;}return -1;}
};

find解答

class Solution {
public:int strStr(string haystack, string needle) {return haystack.find(needle);}
};

暴力

class Solution {
public:int strStr(string haystack, string needle) {// 时间复杂度O(MN)，空间复杂度O(1)int n1 = haystack.size(), n2 = needle.size(), j;for (int i = 0; i < n1 - n2 + 1; ++i) {if (haystack[i] != needle[0]) continue;for (j = 1; j < n2; ++j) {if (needle[j] != haystack[i + j]) break;}if (j == n2) return i;}return -1;}
};