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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
设状态为f[i][j]，表示A[0,i] 和B[0,j] 之间的最小编辑距离。设A[0,i] 的形式是str1c，B[0,j] 的形式是str2d，
1. 如果c==d，则f[i][j]=f[i-1][j-1]；
2. 如果c!=d，
(a) 如果将c 替换成d，则f[i][j]=f[i-1][j-1]+1；
(b) 如果在c 后面添加一个d，则f[i][j]=f[i][j-1]+1；
(c) 如果将c 删除，则f[i][j]=f[i-1][j]+1；
源码：Java版本
算法分析：二维动态规划。时间复杂度O(m*n)，空间复杂度O(m*n)
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public class Solution {
public int minDistance(String word1, String word2) {
int m=word1.length();
int n=word2.length();
int[][] f=new int[m+1][n+1];
for(int i=0;i
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