dp方程“
1、初始化;dp[0][i]=i; dp[j][0]=j;
2.dp[i][j]= dp[i-1][j-1](相等)
dp[i-1][j]+1 ,,dp[i][j-1]+1; dp[i-1][j-1] (这个对应是改的况)
注意字符串下标开始位置就OK了
1 public class Solution { 2 public int minDistance(String word1, String word2) { 3 char c1[]=word1.toCharArray(); 4 char c2[]=word2.toCharArray(); 5 int len1=word1.length(); 6 int len2=word2.length(); 7 int dp[][]=new int[len1+1][len2+1]; 8 int i,j; 9 for(i=0;i<len2+1;i++) dp[0][i]=i; 10 for(j=0;j<len1+1;j++) dp[j][0]=j; 11 for(i=1;i<len1+1;i++) 12 { 13 14 for( j=1;j<len2+1;j++) 15 { 16 if(c1[i-1]==c2[j-1]) dp[i][j]=dp[i-1][j-1]; 17 else{ 18 dp[i][j]=Math.min(dp[i-1][j],dp[i][j-1])+1; 19 20 dp[i][j]=Math.min(dp[i][j],dp[i-1][j-1]+1); 21 22 23 24 } 25 26 27 28 29 } 30 31 32 33 } 34 35 36 return dp[len1][len2]; 37 38 39 } 40 }