[LeetCode] 28. Implement strStr() 实现strStr()函数

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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