Leetcode72. Edit Distance

题目来源：Leetcode

Given two words word1 and word2, find the minimum number of steps required to convertword1 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

思路：

将两个字符串分别构成行和列 * dp[i][j]表示word1(0...i)转换为word（0.。。j）所需要的步数

我自己的代码通过率：1143/1146

class Solution {    public int minDistance(String word1, String word2) {        int len1=word1.length();        int len2=word2.length();        if(word1.length()==0)            return word2.length();        if (word2.length()==0)            return word1.length();        int dp[][]=new int[len1][len2];        if (word1.charAt(0)==word2.charAt(0))            dp[0][0]=0;        else            dp[0][0]=1;        for (int i=1;i<len1;i++){            if (word1.charAt(i)==word2.charAt(0))                dp[i][0]=dp[i-1][0];            else                dp[i][0]=dp[i-1][0]+1;        }        for (int i=1;i<len2;i++){            if (word1.charAt(0)==word2.charAt(i))                dp[0][i]=dp[0][i-1];            else                dp[0][i]=dp[0][i-1]+1;        }        for (int i=1;i<len1;i++){            for (int j=1;j<len2;j++){                if (word1.charAt(i)==word2.charAt(j))                    dp[i][j]=dp[i-1][j-1];                else {                    //if(i==j)                     //   dp[i][j]=Math.min(dp[i-1][j-1],dp[i][j-1])+1;                    dp[i][j]=Math.min(Math.min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1;            }            }        }        return dp[len1-1][len2-1];    }}

修正：

public int minDistance(String word1, String word2) {        int len1=word1.length()+1;        int len2=word2.length()+1;        if(word1.length()==0)            return word2.length();        if (word2.length()==0)            return word1.length();        int dp[][]=new int[len1][len2];        for (int i = 0; i < len1; i++)            dp[i][0] = i;        for (int i = 0; i < len2; i++)            dp[0][i] = i;        for (int i=1;i<len1;i++){            for (int j=1;j<len2;j++){                if (word1.charAt(i-1)==word2.charAt(j-1))                    dp[i][j]=dp[i-1][j-1];                else {                    //if(i==j)                     //   dp[i][j]=Math.min(dp[i-1][j-1],dp[i][j-1])+1;                    dp[i][j]=Math.min(Math.min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1])+1;            }            }        }        return dp[len1-1][len2-1];    }