LeetCode: 28. Implement strStr()

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28. Implement strStr()

LeetCode - Implement strStr()

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

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 strstr() and Java’s indexOf().

解题分析

直接截取字符串然后判断相等，算不算作弊？😆

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public int strStr(String haystack, String needle) {
  if (Objects.isNull(needle) || needle.length() == 0) {
    return 0;
  }
  for (int i = 0; i <= haystack.length() - needle.length(); i++) {
    if (haystack.charAt(i) == needle.charAt(0)) {
      String sub = haystack.substring(i, i + needle.length());
      if (needle.equals(sub)) {
        return i;
      }
    }
  }
  return -1;
}

// TODO 有必要学一下 KMP 算法了啊！
public int strStr1(String source, String target) {
  if (source == null || target == null || target.length() > source.length()) {
    return -1;
  }
  if (target.length() == 0) {
    return 0;
  }
  char[] sA = source.toCharArray();
  char[] tA = target.toCharArray();
  for (int i = 0; i < source.length(); i++) {
    if (contain(sA, i, tA)) {
      return i;
    }
  }
  return -1;
}

private boolean contain(char[] source, int index, char[] target) {
  if (source.length - index < target.length) {
    return false;
  }
  for (int i = index, j = 0; j < target.length; i++, j++) {
    if (source[i] != target[j]) {
      return false;
    }
  }
  return true;
}