# [LeetCode] 583. Delete Operation for Two Strings 两个字符串的删除操做

2021年09月15日 阅读数：1

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.html

Example 1:java

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

Note:python

1. The length of given words won't exceed 500.
2. Characters in given words can only be lower-case letters.

Java1:htm

```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));
}
}
```

Java2:blog

```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()];
}
}```

Java:递归

```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()];
}
}　　```

Python1:字符串

```class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
return len(word1) + len(word2) - 2 * self.lcs(word1, word2)

def lcs(self, word1, word2):
len1, len2 = len(word1), len(word2)
dp = [[0] * (len2 + 1) for x in range(len1 + 1)]
for x in range(len1):
for y in range(len2):
dp[x + 1][y + 1] = max(dp[x][y + 1], dp[x + 1][y])
if word1[x] == word2[y]:
dp[x + 1][y + 1] = dp[x][y] + 1
return dp[len1][len2]　　```

Python2:

```class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
m, n = len(word1), len(word2)
dp = [[0] * (n+1) for _ in xrange(2)]
for i in xrange(m):
for j in xrange(n):
dp[(i+1)%2][j+1] = max(dp[i%2][j+1], \
dp[(i+1)%2][j], \
dp[i%2][j] + (word1[i] == word2[j]))
return m + n - 2*dp[m%2][n]
```

C++:

```class Solution {
public:
int minDistance(string word1, string word2) {
int n1 = word1.size(), n2 = word2.size();
vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));
for (int i = 1; i <= n1; ++i) {
for (int j = 1; j <= n2; ++j) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return n1 + n2 - 2 * dp[n1][n2];
}
};
```

C++:

```class Solution {
public:
int minDistance(string word1, string word2) {
int n1 = word1.size(), n2 = word2.size();
vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, 0));
for (int i = 0; i <= n1; ++i) dp[i][0] = i;
for (int j = 0; j <= n2; ++j) dp[0][j] = j;
for (int i = 1; i <= n1; ++i) {
for (int j = 1; j <= n2; ++j) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[n1][n2];
}
};
```

[LeetCode] 712. Minimum ASCII Delete Sum for Two Strings 两个字符串的最小ASCII删除和