Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

在一个字符串中找另一个字符串第一次出现的位置。

解法2：从第1个字母开始循环，检查由当前字母s[i]到s[i+len(needle)]是否等于needle。T: O(n * k), S: O(k)

解法2：Knuth-Morris-Pratt字符串查找算法（简称为KMP算法）可在一个主文本字符串S内查找一个词W的出现位置。T: O(n + k), S: O(k)

Java:

public int strStr(String haystack, String needle) {
for (int i = 0; ; i++) {
for (int j = 0; ; j++) {
if (j == needle.length()) return i;
if (i + j == haystack.length()) return -1;
if (needle.charAt(j) != haystack.charAt(i + j)) break;
}
}
}

Java:

public class Solution {
public int strStr(String haystack, String needle) {
int l1 = haystack.length(), l2 = needle.length();
if (l1 < l2) {
return -1;
} else if (l2 == 0) {
return 0;
}
int threshold = l1 - l2;
for (int i = 0; i <= threshold; ++i) {
if (haystack.substring(i,i+l2).equals(needle)) {
return i;
}
}
return -1;
}
}　　

Python: T: O(n*k)

class Solution(object):
def strStr(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
for i in xrange(len(haystack) - len(needle) + 1):
if haystack[i : i + len(needle)] == needle:
return i
return -1

Python: KMP, T: O(n + k)

class Solution(object):
def strStr(self, haystack, needle):
"""
:type haystack: str
:type needle: str
:rtype: int
"""
if not needle:
return 0

return self.KMP(haystack, needle)

def KMP(self, text, pattern):
prefix = self.getPrefix(pattern)
j = -1
for i in xrange(len(text)):
while j > -1 and pattern[j + 1] != text[i]:
j = prefix[j]
if pattern[j + 1] == text[i]:
j += 1
if j == len(pattern) - 1:
return i - j
return -1

def getPrefix(self, pattern):
prefix = [-1] * len(pattern)
j = -1
for i in xrange(1, len(pattern)):
while j > -1 and pattern[j + 1] != pattern[i]:
j = prefix[j]
if pattern[j + 1] == pattern[i]:
j += 1
prefix[i] = j
return prefix　　

C++:

class Solution {
public:
int strStr(string haystack, string needle) {
if (needle.empty()) return 0;
int m = haystack.size(), n = needle.size();
if (m < n) return -1;
for (int i = 0; i <= m - n; ++i) {
int j = 0;
for (j = 0; j < n; ++j) {
if (haystack[i + j] != needle[j]) break;
}
if (j == n) return i;
}
return -1;
}
};

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