*LeetCode-Edit Distance

常规做法是用一个mn matrix 然后update条件是假如这一位相等 那么就等于左上角 假如不等就是左，上， 左上三个的min加一

初始化要多处一行一列 初始化成和index相同

然后优化算法是只用一行的array加上一个数字来记录

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

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