Edit Distance || 计算字符串相似度

标签：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][n];
    }
}

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];
    }
}

Edit Distance || 计算字符串相似度,布布扣,bubuko.com

Edit Distance || 计算字符串相似度

标签：算法   dp   

原文地址：http://blog.csdn.net/dutsoft/article/details/38661291