# 583.Delete Operation for Two Strings

## Description

Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".


Note:

The length of given words won’t exceed 500.

Characters in given words can only be lower-case letters.

cnblog和belong，最长公共子序列是blog，而最长公共子串是lo

public class Solution {
public int minDistance(String s1, String s2) {
return s1.length() + s2.length() - 2 * lcs(s1, s2, s1.length(), s2.length());
}
public int lcs(String s1, String s2, int m, int n) {
if (m == 0 || n == 0)
return 0;
if (s1.charAt(m - 1) == s2.charAt(n - 1))
return 1 + lcs(s1, s2, m - 1, n - 1);
else
return Math.max(lcs(s1, s2, m, n - 1), lcs(s1, s2, m - 1, n));
}
}


class Solution {
public int minDistance(String word1, String word2) {
int dp[][]=new int[word1.length()+1][word2.length()+1];
for(int i=0;i<word1.length()+1;++i){
for(int j=0;j<word2.length()+1;++j){
if(i==0||j==0)
continue;
if(word1.charAt(i-1)==word2.charAt(j-1))
dp[i][j]=dp[i-1][j-1]+1;
else
dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);
}
}
return word1.length()+word2.length()-2*dp[word1.length()][word2.length()];
}
}


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


### 学籍管理系统文档

#### 北邮教务系统评教脚本

Published on September 17, 2017

#### 72.Edit distance

Published on September 17, 2017