题目链接

Python3

直觉解法

class Solution:def strStr(self, haystack: str, needle: str) -> int:pn, ph = 0, 0n = len(needle) h = len(haystack)while ph < h:if haystack[ph] == needle[pn]:if pn == n-1: # 1234   123return ph - len(needle) + 1else: pn += 1ph += 1else: ## 1234  122ph = ph - pn + 1pn = 0return -1

方法一： 暴力解法 $⟮O(n\times m)、O(1)⟯$

class Solution:def strStr(self, haystack: str, needle: str) -> int:for i in range(0, len(haystack)-len(needle)+1):if haystack[i:i+len(needle)] == needle:return i return -1

⭐ 方法二：Knuth-Morris-Pratt 算法 [KMP算法] $⟮O(m+n)、O(m)⟯$ 【空间换时间】

从而实现 更快地 跳转

参考链接1
参考链接2： KMP字符串匹配算法2

官方解法链接

class Solution:def strStr(self, haystack: str, needle: str) -> int:h = len(haystack)n = len(needle)# 获取 needle 的前缀信息lis = [0] * n j = 0  # 前后  子串长度for i in range(1, n): # 需要前后比较， 1个字符无法比较，因此从 i=1 开始while j > 0 and needle[i] != needle[j]: j = lis[j-1] # 和之前 相等的长度一样if needle[i] == needle[j]:j += 1  lis[i] = j # 0  1 2 3 4 5 6# a  a b a a a b   needle# 0  1 0 1 2 2 3  前缀子串 后缀子串 相同的长度  若是 needle[j] 不匹配了，注意是 lis[j-1] 存储的才是 待比较的下一个 j# a  a c a a# 根据上述的 信息进行 匹配j = 0  # 遍历 needle 下标for i in range(0, h): # 遍历 haystack 下标while j > 0 and haystack[i] != needle[j]:  #               j = lis[j-1]  # 这里 根据 前后缀 信息  快速跳转  if haystack[i] == needle[j]:j += 1if j == n: # needle 已全部匹配 因为前面的if 成立，j多加了1 return i - n + 1return -1 

链接

class Solution:def strStr(self, haystack: str, needle: str) -> int:def get_next():for i in range(1, len(needle)):k = fπ[i-1]while needle[i] != needle[k]:if k == 0:k -= 1break else:k = fπ[k-1]fπ[i] = k+1n = len(needle)fπ = [0] * n get_next()   # 生成 needle 的next 数组i = 0 j = 0 while i < len(haystack):if haystack[i] == needle[j]:i += 1j += 1elif j == 0:i += 1else:j = fπ[j-1]if j >= n:return i - n return -1

C++

KMP 算法 $⟮O(h+n)、O(n)⟯$

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

暴力解法

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