Difficulty: Easy
Implement .
Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
Example 1:
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| Input: haystack = "hello", needle = "ll"
Output: 2
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Example 2:
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| Input: haystack = "aaaaa", needle = "bba"
Output: -1
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Clarification:
What should we return when needle is an empty string? This is a great question to ask during an interview.
For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C’s and Java’s .
Solution
Language: Java
双重循环
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| class Solution {
public int strStr(String haystack, String needle) {
if (haystack == null || needle == null) {
return -1;
}
int j;
for (int i = 0; i < haystack.length() - needle.length() + 1; i++) {
for (j = 0; j < needle.length(); j++) {
if (haystack.charAt(i + j) != needle.charAt(j)) {
break;
}
}
if (j == needle.length()) {
return i;
}
}
return -1;
}
}
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Robin Carp算法
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| class Solution {
public int strStr(String haystack, String needle) {
if (haystack == null || needle == null || haystack.length() < needle.length()) {
return -1;
}
if (needle.length() == 0) {
return 0;
}
int len = needle.length();
int MAX = 1000000;
int MOD = 31;
int target = 0;
int cur = 0;
int HIGH = 1;
for (int i = 0; i < len; i++) {
target = ((target * 31) % MOD + needle.charAt(i)) % MOD;
HIGH = (HIGH * 31) % MOD;
}
for (int i = 0; i < haystack.length(); i++) {
cur = ((cur * 31) % MOD + haystack.charAt(i)) % MOD;
if (i < len - 1) {
continue;
}
if (cur == target && haystack.substring(i - len + 1, i + 1).equals(needle)) {
return i - len + 1;
}
cur = (cur - (haystack.charAt(i - len + 1) * HIGH) % MOD) % MOD;
}
return -1;
}
}
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