Leetcode : Edit Distance

URL https://leetcode.com/problems/edit-distance/description/

题目大意：

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

解题思路

本题思路是 dp[i,j] 表示word1(1,i)转化为word2(1,j)所用最小操作次数。则有下面的几种情况：
1. word1[i]==word2[j] 则 dp(i,j) = dp(i-1,j-1) 即可。
2. word1[i]!=word2[j] 则我们有三种操作，
2.1 替换word1[i] ，dp(i,j) = dp(i-1,j-1)+1
2.2 删除操作word1[i] ，dp(i,j) = dp(i-1,j)+1
2.3 增加到word1尾部 ，dp(i,j) = dp(i,j-1)+1

下面给出代码

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

我们其实可以用一个一维数组搞定。

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