Edit Distance || 计算字符串相似度

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character

b) Delete a character

c) Replace a character
字符串相似度定义为 1/距离

二维dp，O(m*n)空间

public class Solution {
public int minDistance(String word1, String word2) {
int m=word1.length();
int n=word2.length();
int [][]f=new int[m+1][n+1];
for(int i=0;i<=n;i++) f[0][i]=i;
for(int i=0;i<=m;i++) f[i][0]=i;
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(word1.charAt(i-1)==word2.charAt(j-1)){
f[i][j]=f[i-1][j-1];
}
else{
int tmp=Math.min(f[i-1][j],f[i][j-1]);
f[i][j]=Math.min(tmp,f[i-1][j-1])+1;
}
}
}
return f[m]
;
}
}

二维dp，滚动数组，O(n)空间

public class Solution {
public int minDistance(String word1, String word2) {
int m=word1.length();
int n=word2.length();
int []f=new int[n+1];
int upleft=0;
for(int i=0;i<=n;i++) f[i]=i;
for(int i=1;i<=m;i++){
upleft=f[0];
f[0]=i;
for(int j=1;j<=n;j++){
int up=f[j];
if(word1.charAt(i-1)==word2.charAt(j-1)){
f[j]=upleft;
}
else{
int tmp=Math.min(f[j],f[j-1]);
f[j]=Math.min(upleft,tmp)+1;
}
upleft=up;
}
}
return f
;
}
}