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LANG-1277: StringUtils#getLevenshteinDistance reduce memory consumption (closes #189)

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yufcuy 2016-09-13 17:38:00 +08:00 committed by pascalschumacher
parent 3433a94e25
commit 103b64a373
1 changed files with 13 additions and 35 deletions
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48
src/main/java/org/apache/commons/lang3/StringUtils.java
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@ -7737,15 +7737,11 @@ public class StringUtils {
* insertion or substitution).</p>
*
* <p>The previous implementation of the Levenshtein distance algorithm
* was from <a href="https://web.archive.org/web/20120604192456/http://www.merriampark.com/ld.htm">
* https://web.archive.org/web/20120604192456/http://www.merriampark.com/ld.htm</a></p>
*
* <p>Chas Emerick has written an implementation in Java, which avoids an OutOfMemoryError
* which can occur when my Java implementation is used with very large strings.<br>
* This implementation of the Levenshtein distance algorithm
* is from <a href="https://web.archive.org/web/20120526085419/http://www.merriampark.com/ldjava.htm">
* was from <a href="https://web.archive.org/web/20120526085419/http://www.merriampark.com/ldjava.htm">
* https://web.archive.org/web/20120526085419/http://www.merriampark.com/ldjava.htm</a></p>
*
* <p>This implementation only need one single-dimensional arrays of length s.length() + 1</p>
*
* <pre>
* StringUtils.getLevenshteinDistance(null, *) = IllegalArgumentException
* StringUtils.getLevenshteinDistance(*, null) = IllegalArgumentException
@ -7773,20 +7769,8 @@ public class StringUtils {
}
/*
The difference between this impl. and the previous is that, rather
than creating and retaining a matrix of size s.length() + 1 by t.length() + 1,
we maintain two single-dimensional arrays of length s.length() + 1. The first, d,
is the 'current working' distance array that maintains the newest distance cost
counts as we iterate through the characters of String s. Each time we increment
the index of String t we are comparing, d is copied to p, the second int[]. Doing so
allows us to retain the previous cost counts as required by the algorithm (taking
the minimum of the cost count to the left, up one, and diagonally up and to the left
of the current cost count being calculated). (Note that the arrays aren't really
copied anymore, just switched...this is clearly much better than cloning an array
or doing a System.arraycopy() each time through the outer loop.)
Effectively, the difference between the two implementations is this one does not
cause an out of memory condition when calculating the LD over two very large strings.
This implementation use two variable to record the previous cost counts,
So this implementation use less memory than previous impl.
*/
int n = s.length(); // length of s
@ -7807,16 +7791,14 @@ public class StringUtils {
m = t.length();
}
int p[] = new int[n + 1]; //'previous' cost array, horizontally
int d[] = new int[n + 1]; // cost array, horizontally
int _d[]; //placeholder to assist in swapping p and d
int p[] = new int[n + 1];
// indexes into strings s and t
int i; // iterates through s
int j; // iterates through t
int upper_left;
int upper;
char t_j; // jth character of t
int cost; // cost
for (i = 0; i <= n; i++) {
@ -7824,23 +7806,19 @@ public class StringUtils {
}
for (j = 1; j <= m; j++) {
upper_left = p[0];
t_j = t.charAt(j - 1);
d[0] = j;
p[0] = j;
for (i = 1; i <= n; i++) {
upper = p[i];
cost = s.charAt(i - 1) == t_j ? 0 : 1;
// minimum of cell to the left+1, to the top+1, diagonally left and up +cost
d[i] = Math.min(Math.min(d[i - 1] + 1, p[i] + 1), p[i - 1] + cost);
p[i] = Math.min(Math.min(p[i - 1] + 1, p[i] + 1), upper_left + cost);
upper_left = upper;
}
// copy current distance counts to 'previous row' distance counts
_d = p;
p = d;
d = _d;
}
// our last action in the above loop was to switch d and p, so p now
// actually has the most recent cost counts
return p[n];
}