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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
StringUtils.getLevenshteinDistance("hello", "hallo") = 1Java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;
}
}
return d[n];
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