Java滚动数组计算编辑距离操作示例

Java滚动数组计算编辑距离操作示例

本文实例讲述了java滚动数组计算编辑距离操作。分享给大家供大家参考，具体如下：

编辑距离（Edit Distance），也称Levenshtein距离，是指由一个字符串转换为另一个字符串所需的最少编辑次数。

下面的代码摘自org.apache.commons.lang.StringUtils

用法示例：

StringUtils.getLevenshteinDistance(null, *) = IllegalArgumentException

StringUtils.getLevenshteinDistance(*, null) = IllegalArgumentException

StringUtils.getLevenshteinDistance("","") = 0

StringUtils.getLevenshteinDistance("","a") = 1

StringUtils.getLevenshteinDistance("aaapppp", "") = 7

StringUtils.getLevenshteinDistance("frog", "fog") = 1

StringUtils.getLevenshteinDistance("fly", "ant") = 3

StringUtils.getLevenshteinDistance("elephant", "hippo") = 7

StringUtils.getLevenshteinDistance("hippo", "elephant") = 7

StringUtils.getLevenshteinDistance("hippo", "zzzzzzzz") = 8

StringUtilhttp://s.getLevenshteinDistance("hello", "hallo") = 1

Java代码：

public static int getLevenshteinDistance(String s, String t) {

if (s == null || t == null) {

throw new IllegalArgumentException("Strings must not be null");

}

int n = s.length(); // length of s

int m = t.length(); // length of t

if (n == 0) {

return m;

} else if (m == 0) {

return n;

}

if (n > m) {

// swap the input strings to consume less memory

String tmp = s;

s = t;

t = tmp;

n = m;

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

// indexes into strings s and t

int i; // iterates through s

int j; // iterates through t

char t_j; // jth character of t

int cost; // cost

for (i = 0; i<=n; i++) {

p[i] = i;

}

for (j = 1; j<=m; j++) {

t_j = t.charAt(j-1);

d[0] = j;

for (i=1; i<=n; 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);

}

// 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];

}

实际上，上述代码的空间复杂度还可以进一步简化，使用一维数组替换滚动数组。

Java代码：

public int minDistance(String s, String t) {

if (s == null || t == null) {

throw new IllegalArgumentException("Strings must not be null");

}

int n = s.length(); // length of s

int m = t.length(); // length of t

if (n == 0) {

return m;

} else if (m == 0) {

return n;

}

if (n > m) {

// swap the input strings to consume less memory

String tmp = s;

s = t;

t = tmp;

n = m;

m = t.length();

}

int d[] = new int[n+1]; // cost array, horizontally

// indexes into strings s and t

int i; // iterates through s

int j; // iterates through t

char t_j; // jth character of t

int cost; // cost

for (i = 0; i<=n; i++) {

d[i] = i;

}

for (j = 1; j<=m; j++) {

t_j = t.charAt(j-1);

int pre = d[0];

d[0] = j;

for (i=1; i<=n; i++) {

int temp = d[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, d[i]+1), pre+cost);

pre = temp;

}

}http://

return d[n];

}

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