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LeetCode 72 Edit Distance
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
思路:本题属于典型的动态规划问题,动态规划+备忘录方法,使用distance[i][j]来存储就是取word1前i个character和word2前j个character所得的最短编辑距离,并使用.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); }
LeetCode 72 Edit Distance
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