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leetcode 28. implement strstr()

admin 11月 12, 2020 0

28. Implement strStr()

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

Example 2:

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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 and Java’s .

Solution

Language: Java
双重循环

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class  {
      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;
     }
}

Robin Carp算法

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class  {
    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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