前言

KMP真厉害，刷题刷到 28.找出字符串中第一个匹配项的下标 和 1668.最大重复子字符串

next 数组用来匹配不上时，前缀 $j$ 可以快速回退到 $next[j-1]$ 的位置。

void getNext(vector<int>& next, const string& s) {int j = 0;  // 后缀// next[0] = 0;  // next初始化为 0for (int i = 1; i < s.size(); i++) {while (j > 0 && s[i] != s[j])j = next[j-1];  // 往回跳if (s[i] == s[j])j++;next[i] = j;   // 更新next数组}
}

题目

28 找出字符串中第一个匹配项的下标

模版题， getNext可以重复使用

class Solution {
public:void getNext(vector<int>& next, const string& s) {int j = 0;  // 后缀// next[0] = 0;  // next初始化为 0for (int i = 1; i < s.size(); i++) {while (j > 0 && s[i] != s[j])j = next[j-1];  // 往回跳if (s[i] == s[j])j++;next[i] = j;   // 更新next数组}}int strStr(string haystack, string needle) {// needle是子串if (haystack.size() == 0 || needle.size() == 0)return -1;vector<int> next(needle.size(), 0);// 构建 next 数组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 - j + 1;}return -1;}
};

1668 最大重复子字符串

KMP + 动态规划，很难想象这是 简单 题

class Solution {
public:void getNext(vector<int>& next, string s) {int j = 0;              // 后缀for (int i = 1; i < s.size(); i++) {while (j > 0 && s[i] != s[j])j = next[j - 1];  // 回退if (s[i] == s[j])j++;next[i] = j;}}int maxRepeating(string sequence, string word) {int n = sequence.size(), m = word.size();if (n < m)return 0;vector<int> next(m, 0);getNext(next, word);// 动态规划 dp[i]表示 sequence[0..i]最多有dp[i]个wordvector<int> dp(n, 0); int j = 0;for (int i = 0; i < sequence.size(); i++) {while(j > 0 && sequence[i] != word[j])j = next[j - 1];if (sequence[i] == word[j])j++;if (j == m) {  // 匹配上了开始 dp 操作if (i >= m)dp[i] = dp[i - m] + 1;elsedp[i] = 1;}}int maxVal = 0;for (int val: dp)maxVal = max(maxVal, val);return maxVal;}
};