Edit Distance || 計算字串相似性

標籤：演算法   dp   

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

字串相似度定義為 1/距離

二維dp，O(m*n)空間

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][n];    }}

二維dp，滾動數組，O(n)空間

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[n];    }}