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LeetCode Implement strStr(kmp或者BM)

2016-07-27 18:17 302 查看
题意:求一字符串在另一字符串的中的索引

思路:kmp

代码如下:

public class Solution {
private int[] kmp_table(String s)
{
int[] next = new int[s.length()];

next[0] = -1;

for (int i = 1; i < s.length(); i++)
{
int j = next[i - 1];
while (j > -1 && s.charAt(j) != s.charAt(i - 1))
{
j = next[j];
}

next[i] = j + 1;
}

return next;
}

private int kmp_search(String text, String pattern)
{
int[] next = kmp_table(pattern);
int match_start = 0, pattern_start = 0;

while (match_start + pattern_start < text.length())
{
while (match_start + pattern_start < text.length() && text.charAt(match_start + pattern_start) == pattern.charAt(pattern_start))
{
if (++pattern_start == pattern.length()) return match_start;
}

match_start += pattern_start - next[pattern_start];
pattern_start = next[pattern_start] > -1 ? next[pattern_start] : 0;
}

return -1;
}

public int strStr(String haystack, String needle)
{
if (haystack.compareTo(needle) == 0) return 0;
if (needle.isEmpty()) return 0;
if (haystack.length() < needle.length()) return -1;

return kmp_search(haystack, needle);
}
}

解法二

public class Solution
{
private int[] kmp_table(String s)
{
int[] next = new int[s.length()];

next[0] = -1;
int j = -1, i = 0;

while (i < s.length())
{
while (j > -1 && s.charAt(j) != s.charAt(i))
{
j = next[j];
}

j++;
i++;
if (i < s.length() && j < s.length())
{
if (s.charAt(i) == s.charAt(j))
{
next[i] = next[j];
}
else
{
next[i] = j;
}
}
}

return next;
}

private int kmp_search(String text, String pattern)
{
int[] next = kmp_table(pattern);
int match_start = 0, pattern_start = 0;

while (match_start < text.length())
{
while (pattern_start > -1 && text.charAt(match_start) != pattern.charAt(pattern_start))
{
pattern_start = next[pattern_start];
}

pattern_start++;
match_start++;
if (pattern_start == pattern.length())
{
return match_start - pattern_start;
}
}

return -1;
}

public int strStr(String haystack, String needle)
{
if (haystack.compareTo(needle) == 0) return 0;
if (needle.isEmpty()) return 0;
if (haystack.length() < needle.length()) return -1;

return kmp_search(haystack, needle);
}
}


解法三:bm算法

public class Solution
{
private Map<Character, Integer> skip_;
private int[] suffix_;

private void build_skip_table(String s)
{
skip_ = new HashMap<>();
for (int i = 0; i < s.length(); i++)
{
skip_.put(s.charAt(i), i);
}
}

private int[] compute_bm_prefix(String s)
{
if (s.length() == 0) return null;

int[] prefix = new int[s.length()];
prefix[0] = 0;
int k = 0;
for (int i = 1; i < s.length(); i++)
{
while (k > 0 && s.charAt(k) != s.charAt(i))
{
k = prefix[k - 1];
}

if (s.charAt(k) == s.charAt(i)) k++;
prefix[i] = k;
}

return prefix;
}

private void build_suffix_table(String s)
{
if (s.length() == 0) return ;

int[] prefix = compute_bm_prefix(s);

String reverse_s = new StringBuilder(s).reverse().toString();
int[]  reverse_prefix = compute_bm_prefix(reverse_s);

suffix_ = new int[s.length() + 1];
for (int i = 0; i <= s.length(); i++)
{
suffix_[i] = s.length() - prefix[s.length() - 1];
}

for (int i = 0; i < reverse_s.length(); i++)
{
int index = reverse_s.length() - reverse_prefix[i];
int shift = i - reverse_prefix[i] + 1;
if (suffix_[index] > shift)
{
suffix_[index] = shift;
}
}
}

private int do_search(String text, String pattern)
{
build_skip_table(pattern);
build_suffix_table(pattern);

int index_end = text.length() - pattern.length();
int i = 0;
while (i <= index_end)
{
int j = pattern.length();
while (text.charAt(i +  j - 1) == pattern.charAt(j - 1))
{
j--;
if (j == 0) return i;
}

int k = skip_.containsKey(text.charAt(j - 1)) ? skip_.get(pattern.charAt(j - 1)) : -1;
int m = j - 1 - k;
if (k < j && m > suffix_[j])
{
i += m;
}
else
{
i += suffix_[j];
}
}

return -1;
}

public int strStr(String haystack, String needle)
{
if (haystack.compareTo(needle) == 0) return 0;
if (needle.isEmpty()) return 0;
if (haystack.length() < needle.length()) return -1;

return do_search(haystack, needle);
}
}
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