关键在于如何求next数组
void getNext(int *next, const string &s) { int j = -1; next[0] = j; for (int i = 1; i < s.size(); i++) { // next[j + 1]指向匹配好的前缀的下一个字符 // i指向后缀末尾位置 while (j >= 0 && s[i] != s[j + 1]) { j = next[j]; } if (s[i] == s[j + 1]) { j++;//j+1表示最长相等子串的长度,所以字符相等时,长度递增 } next[i] = j; } }
#include <string> using std::string; class Solution { public: void getNext(int *next, const string &s) { int j = -1; next[0] = j; for (int i = 1; i < s.size(); i++) { // next[j + 1]指向匹配好的前缀的下一个字符 // i指向后缀末尾位置 while (j >= 0 && s[i] != s[j + 1]) { j = next[j]; } if (s[i] == s[j + 1]) { 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 = -1; for (int i = 0; i < haystack.size(); i++) { //出现不等时,跳到最长相等子串的下一个字符处 while (j >= 0 && haystack[i] != needle[j + 1]) { j = next[j]; } if (haystack[i] == needle[j + 1]) { j++; } if (j == (needle.size() - 1)) return i - needle.size() + 1; } return -1; } };