leetcode 72. Edit Distance（动态规划）
Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
Insert a character
Delete a character
Replace a character
Example 1:
Input: word1 = “horse”, word2 = “ros”
Output: 3
Explanation:
horse -> rorse (replace ‘h’ with ‘r’)
rorse -> rose (remove ‘r’)
rose -> ros (remove ‘e’)
Example 2:
Input: word1 = “intention”, word2 = “execution”
Output: 5
Explanation:
intention -> inention (remove ‘t’)
inention -> enention (replace ‘i’ with ‘e’)
enention -> exention (replace ‘n’ with ‘x’)
exention -> exection (replace ‘n’ with ‘c’)
exection -> execution (insert ‘u’)
- 动态规划
C++
// O(n^2) space
class Solution {
public:
int minDistance(string a, string b) {
int m=a.size(),n=b.size();
vector<vector<int>> f(m+1,vector<int>(n+1,0));
for(int i=1;i<=m;i++)f[i][0]=i; // a[0...i) -> b[0...0) needs i operation
for(int i=1;i<=n;i++)f[0][i]=i; // a[0,0) -> b[0...i) needs i operation
for(int i=1;i<=m;i++)
for(int j=1;j<=n;j++){
if(a[i-1]==b[j-1])f[i][j]=f[i-1][j-1];
else f[i][j]=min(f[i-1][j-1],min(f[i][j-1],f[i-1][j]))+1; // replace,delete or insert
}
return f[m][n];
}
};
// two vectors O(2*n) space
class Solution {
public:
int minDistance(string a, string b) {
int m=a.size(),n=b.size();
vector<int> pre(n+1,0),cur(n+1,0);
for(int i=1;i<=n;i++)pre[i]=i;
for(int i=1;i<=m;i++){
cur[0]=i;
for(int j=1;j<=n;j++){
if(a[i-1]==b[j-1])cur[j]=pre[j-1];
else cur[j]=min(pre[j-1],min(pre[j],cur[j-1]))+1;
}
fill(pre.begin(),pre.end(),0);
swap(pre,cur);
}
return pre[n];
}
};
// one vector O(n) space
class Solution {
public:
int minDistance(string a, string b) {
int m=a.size(),n=b.size(),p;
vector<int> f(n+1,0);
for(int i=1;i<=n;i++)f[i]=i;
for(int i=1;i<=m;i++){
p=f[0];
f[0]=i;
for(int j=1;j<=n;j++){
int tmp=f[j];
if(a[i-1]==b[j-1])f[j]=p;
else f[j]=min(p,min(f[j-1],f[j]))+1;
p=tmp;
}
}
return f[n];
}
};
Java
class Solution {
public int minDistance(String a, String b) {
int m=a.length();
int n=b.length();
int[][] f=new int[m+1][n+1];
for(int i=1;i<=m;i++)
f[i][0]=i;
for(int i=1;i<=n;i++)
f[0][i]=i;
for(int i=1;i<=m;i++)
for(int j=1;j<=n;j++){
if(a.charAt(i-1)==b.charAt(j-1))f[i][j]=f[i-1][j-1];
else{
f[i][j]=Math.min(f[i-1][j-1],Math.min(f[i-1][j],f[i][j-1]))+1;
}
}
return f[m][n];
}
}
class Solution {
public int minDistance(String a, String b) {
int m=a.length();
int n=b.length();
// int[][] f=new int[m+1][n+1];
int[] pre=new int[n+1];
int[] cur=new int[n+1];
for(int i=1;i<=n;i++)
pre[i]=i;
for(int i=1;i<=m;i++){
cur[0]=i;
for(int j=1;j<=n;j++){
if(a.charAt(i-1)==b.charAt(j-1))cur[j]=pre[j-1];
else{
cur[j]=Math.min(pre[j-1],Math.min(pre[j],cur[j-1]))+1;
}
}
for(int k=0;k<=n;k++){
pre[k]=cur[k];
cur[k]=0;
}
}
return pre[n];
}
}
class Solution {
public int minDistance(String a, String b) {
int m=a.length();
int n=b.length();
int p;
// int[][] f=new int[m+1][n+1];
int[] pre=new int[n+1];
// int[] cur=new int[n+1];
for(int i=1;i<=n;i++)
pre[i]=i;
for(int i=1;i<=m;i++){
p=pre[0];
pre[0]=i;
for(int j=1;j<=n;j++){
int tmp=pre[j];
if(a.charAt(i-1)==b.charAt(j-1))pre[j]=p;
else{
pre[j]=Math.min(p,Math.min(pre[j],pre[j-1]))+1;
}
p=tmp;
}
}
return pre[n];
}
}
Python
class Solution:
def minDistance(self, a: str, b: str) -> int:
m,n=len(a),len(b)
f=[[0]*(n+1) for _ in range(m+1)]
for i in range(1,m+1):
f[i][0]=i
for i in range(1,n+1):
f[0][i]=i
for i in range(1,m+1):
for j in range(1,n+1):
if a[i-1]==b[j-1]:
f[i][j]=f[i-1][j-1]
else:
f[i][j]=min(f[i-1][j-1],min(f[i-1][j],f[i][j-1]))+1
return f[m][n]
Go
func minDistance(a string, b string) int {
m,n:=len(a),len(b)
f:=make([][]int, m+1)
for i:=range f{
f[i]=make([]int,n+1)
f[i][0]=i
}
for j:=range f[0]{
f[0][j]=j
}
for i:=1;i<=m;i++ {
for j:=1;j<=n;j++ {
if a[i-1]==b[j-1] {
f[i][j]=f[i-1][j-1]
} else {
f[i][j]=min(f[i-1][j-1],min(f[i-1][j],f[i][j-1]))+1
}
}
}
return f[m][n]
}
func min(a,b int) int {
if a>b {
return b
}
return a
}
