class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();
int[][] dis = new int[m + 1][n + 1];
//初始化
for(int i = 0; i <= m; i++) dis[i][0] = i;
for(int j = 0; j <= n; j++) dis[0][j] = j;
for(int i = 1; i <= m; i++){
for(int j = 1; j <= n; j++){
dis[i][j] = Integer.MAX_VALUE;
if(word1.charAt(i - 1) == word2.charAt(j - 1)){
dis[i][j] = Math.min(dis[i - 1][j] + 1, dis[i][j - 1] + 1);
dis[i][j] = Math.min(dis[i][j], dis[i - 1][j - 1]);
}
else{
dis[i][j] = Math.min(dis[i - 1][j] + 1, dis[i][j - 1] + 1);
dis[i][j] = Math.min(dis[i][j], dis[i - 1][j - 1] + 1);
}
}
}
return dis[m][n];
}
}
归纳:
分为两种情况,word1和word2当前值相同和不同的情况,先初始化,考虑word1和word2为null时,添加元素,初始化完毕,如果word1和word2当前值相同,比如12b到ab有三种情况,(12b=a )+ 1(1表示给a添加一个b),12 = a,(12 = ab) + 1(1表示删除12b中的b),如果word1和word2当前值不相同,比如12c到ab,这时12 = a的情况下需要将c repalce为b,需要加一次操作