LeetCode刷题笔录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

标准Dynamic Programming。思路如下：

两个String看成两个char数组，分别是x[m]以及y
。对于x[1]...x[i]和y[1]...y[j]的edit distance为A[i][j]。那么有如下recursion:

A[i][j]=A[i-1][j-1] if x[i] = y[j];

A[i][j]=min(A[i-1][j], A[i][j-1], A[i-1][j-1])+1 if x[i] != y[j]

其中第二种x[i]!=y[j]的min中的三个元素分别代表insert, delete和replace三种情况，要看哪种最小。

Base case是A[0][j]=j, A[i][0] = i。

code如下：

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