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LeetCode 72 Edit Distance

2024-08-01 02:21:37   223人阅读

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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