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简单 |
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给你两个字符串 haystack 和 needle ，请你在 haystack 字符串中找出 needle 字符串的第一个匹配项的下标（下标从 0 开始）。如果 needle 不是 haystack 的一部分，则返回 -1 。
示例 1：
输入：haystack = "sadbutsad", needle = "sad" 输出：0 解释："sad" 在下标 0 和 6 处匹配。 第一个匹配项的下标是 0 ，所以返回 0 。
示例 2：
输入：haystack = "leetcode", needle = "leeto" 输出：-1 解释："leeto" 没有在 "leetcode" 中出现，所以返回 -1 。
提示：
1 <= haystack.length, needle.length <= 104haystack和needle仅由小写英文字符组成
以字符串 haystack 的每一个字符为起点与字符串 needle 进行比较，若发现能够匹配的索引，直接返回即可。
假设字符串 haystack 长度为 needle 长度为
class Solution:
def strStr(self, haystack: str, needle: str) -> int:
n, m = len(haystack), len(needle)
for i in range(n - m + 1):
if haystack[i : i + m] == needle:
return i
return -1class Solution {
public int strStr(String haystack, String needle) {
if ("".equals(needle)) {
return 0;
}
int len1 = haystack.length();
int len2 = needle.length();
int p = 0;
int q = 0;
while (p < len1) {
if (haystack.charAt(p) == needle.charAt(q)) {
if (len2 == 1) {
return p;
}
++p;
++q;
} else {
p -= q - 1;
q = 0;
}
if (q == len2) {
return p - q;
}
}
return -1;
}
}class Solution {
private:
vector<int> Next(string str) {
vector<int> n(str.length());
n[0] = -1;
int i = 0, pre = -1;
int len = str.length();
while (i < len) {
while (pre >= 0 && str[i] != str[pre])
pre = n[pre];
++i, ++pre;
if (i >= len)
break;
if (str[i] == str[pre])
n[i] = n[pre];
else
n[i] = pre;
}
return n;
}
public:
int strStr(string haystack, string needle) {
if (0 == needle.length())
return 0;
vector<int> n(Next(needle));
int len = haystack.length() - needle.length() + 1;
for (int i = 0; i < len; ++i) {
int j = 0, k = i;
while (j < needle.length() && k < haystack.length()) {
if (haystack[k] != needle[j]) {
if (n[j] >= 0) {
j = n[j];
continue;
} else
break;
}
++k, ++j;
}
if (j >= needle.length())
return k - j;
}
return -1;
}
};func strStr(haystack string, needle string) int {
n, m := len(haystack), len(needle)
for i := 0; i <= n-m; i++ {
if haystack[i:i+m] == needle {
return i
}
}
return -1
}function strStr(haystack: string, needle: string): number {
const m = haystack.length;
const n = needle.length;
for (let i = 0; i <= m - n; i++) {
let isEqual = true;
for (let j = 0; j < n; j++) {
if (haystack[i + j] !== needle[j]) {
isEqual = false;
break;
}
}
if (isEqual) {
return i;
}
}
return -1;
}impl Solution {
pub fn str_str(haystack: String, needle: String) -> i32 {
let haystack = haystack.as_bytes();
let needle = needle.as_bytes();
let m = haystack.len();
let n = needle.len();
let mut next = vec![0; n];
let mut j = 0;
for i in 1..n {
while j > 0 && needle[i] != needle[j] {
j = next[j - 1];
}
if needle[i] == needle[j] {
j += 1;
}
next[i] = j;
}
j = 0;
for i in 0..m {
while j > 0 && haystack[i] != needle[j] {
j = next[j - 1];
}
if haystack[i] == needle[j] {
j += 1;
}
if j == n {
return (i - n + 1) as i32;
}
}
-1
}
}/**
* @param {string} haystack
* @param {string} needle
* @return {number}
*/
var strStr = function (haystack, needle) {
const slen = haystack.length;
const plen = needle.length;
if (slen == plen) {
return haystack == needle ? 0 : -1;
}
for (let i = 0; i <= slen - plen; i++) {
let j;
for (j = 0; j < plen; j++) {
if (haystack[i + j] != needle[j]) {
break;
}
}
if (j == plen) return i;
}
return -1;
};public class Solution {
public int StrStr(string haystack, string needle) {
for (var i = 0; i < haystack.Length - needle.Length + 1; ++i)
{
var j = 0;
for (; j < needle.Length; ++j)
{
if (haystack[i + j] != needle[j]) break;
}
if (j == needle.Length) return i;
}
return -1;
}
}class Solution {
/**
* @param String $haystack
* @param String $needle
* @return Integer
*/
function strStr($haystack, $needle) {
$strNew = str_replace($needle, '+', $haystack);
$cnt = substr_count($strNew, '+');
if ($cnt > 0) {
for ($i = 0; $i < strlen($strNew); $i++) {
if ($strNew[$i] == '+') {
return $i;
}
}
} else {
return -1;
}
}
}Rabin-Karp 算法本质上是利用滑动窗口配合哈希函数对固定长度的字符串哈希之后进行比较，可以将比较两个字符串是否相同的时间复杂度降为
假设字符串 haystack 长度为 needle 长度为
func strStr(haystack string, needle string) int {
n, m := len(haystack), len(needle)
sha, target, left, right, mod := 0, 0, 0, 0, 1<<31-1
multi := 1
for i := 0; i < m; i++ {
target = (target*256%mod + int(needle[i])) % mod
}
for i := 1; i < m; i++ {
multi = multi * 256 % mod
}
for ; right < n; right++ {
sha = (sha*256%mod + int(haystack[right])) % mod
if right-left+1 < m {
continue
}
// 此时 left~right 的长度已经为 needle 的长度 m 了，只需要比对 sha 值与 target 是否一致即可
// 为避免 hash 冲突，还需要确保 haystack[left:right+1] 与 needle 相同
if sha == target && haystack[left:right+1] == needle {
return left
}
// 未匹配成功，left 右移一位
sha = (sha - (int(haystack[left])*multi)%mod + mod) % mod
left++
}
return -1
}function strStr(haystack: string, needle: string): number {
const m = haystack.length;
const n = needle.length;
const next = new Array(n).fill(0);
let j = 0;
for (let i = 1; i < n; i++) {
while (j > 0 && needle[i] !== needle[j]) {
j = next[j - 1];
}
if (needle[i] === needle[j]) {
j++;
}
next[i] = j;
}
j = 0;
for (let i = 0; i < m; i++) {
while (j > 0 && haystack[i] !== needle[j]) {
j = next[j - 1];
}
if (haystack[i] === needle[j]) {
j++;
}
if (j === n) {
return i - n + 1;
}
}
return -1;
}假设字符串 haystack 长度为 needle 长度为