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
.代码如下：public class Solution {
public int minDistance(String word1, String word2) {
int [][]distance=new int [word1.length()+1][word2.length()+1];
for(int i=1;i<=word1.length();distance[i][0]=i,i++);
for(int i=1;i<=word2.length();distance[0][i]=i,i++);
for(int i=1;i<=word1.length();i++){
for(int j=1;j<=word2.length();j++){
if(word1.charAt(i-1)==word2.charAt(j-1)){
distance[i][j]=distance[i-1][j-1];
}else{
distance[i][j]=1+Math.min(Math.min(distance[i-1][j], distance[i][j-1]),distance[i-1][j-1]);
}
}
//System.out.println(Arrays.deepToString(distance));
}
return distance[word1.length()][word2.length()];
}
}后附递归版超时代码： Time Limit Exceeded,Last executed input: "dinitrophenylhydrazine", "acetylphenylhydrazine"
public int minDistance(String word1 ,int index1, String word2,int index2) {
if(index1>=word1.length()&&index2>=word2.length()){
return 0;
}
else if(index1>=word1.length()){
return word2.length()-index2;
}
else if(index2>=word2.length()){
return word1.length()-index1;
}
if(word1.charAt(index1)==word2.charAt(index2)){
return minDistance(word1,index1+1,word2,index2+1);
}else {
int res1=1+Math.min(minDistance(word1,index1+1,word2,index2),minDistance(word1,index1,word2,index2+1));
int res2=1+minDistance(word1,index1+1,word2,index2+1);
return Math.min(res1, res2);
}
}
public int minDistance(String word1, String word2) {
return minDistance(word1 , 0, word2, 0);
}原文地址：http://blog.csdn.net/mlweixiao/article/details/40929845
