题目：
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
题解：
处理这道题也是用动态规划。
动态数组dp[word1.length+1][word2.length+1]
dp[i][j]表示从word1前i个字符转换到word2前j个字符最少的步骤数。
假设word1现在遍历到字符x，word2遍历到字符y（word1当前遍历到的长度为i，word2为j）。
以下两种可能性：
1. x==y，那么不用做任何编辑操作，所以dp[i][j] = dp[i-1][j-1]
2. x != y
(1) 在word1插入y， 那么dp[i][j] = dp[i][j-1] + 1
(2) 在word1删除x， 那么dp[i][j] = dp[i-1][j] + 1
(3) 把word1中的x用y来替换，那么dp[i][j] = dp[i-1][j-1] + 1
最少的步骤就是取这三个中的最小值。
最后返回 dp[word1.length+1][word2.length+1] 即可。
代码如下：
1 public static int minDistance(String word1, String word2) {
2 int len1 = word1.length();
3 int len2 = word2.length();
4
5 // len1+1, len2+1, because finally return dp[len1][len2]
6 int[][] dp = new int[len1 + 1][len2 + 1];
7
8 for (int i = 0; i <= len1; i++)
9 dp[i][0] = i;
for (int j = 0; j <= len2; j++)
dp[0][j] = j;
//iterate though, and check last char
for (int i = 1; i <= len1; i++) {
char c1 = word1.charAt(i-1);
for (int j = 1; j <= len2; j++) {
char c2 = word2.charAt(j-1);
//if last two chars equal
if (c1 == c2) {
//update dp value for +1 length
dp[i][j] = dp[i-1][j-1];
} else {
int replace = dp[i-1][j-1] + 1;
int insert = dp[i-1][j] + 1;
int delete = dp[i][j-1] + 1;
int min = Math.min(replace, insert);
min = Math.min(min,delete);
dp[i][j] = min;
}
}
}
return dp[len1][len2];
}
Reference：
http://www.programcreek.com/2013/12/edit-distance-in-java/
http://blog.****.net/linhuanmars/article/details/24213795