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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:
Example
Given word1 = "mart" and word2 = "karma", return 3.
SOLUTION :
先看问题，minimum number of steps，可以说基本就是DP问题了。
状态：由于是两个字符串，要分别用两个字符串的信息，所以二维DP，f(x,y) = w1前x位 变成 w2的前y位，需要多少步。
方程：情况1:w1的第i位跟w2的第j位相等，那么f(i,j) = f(i - 1, j - 1)
情况2: 不想等：1，insert ，insert后，w2的j位被匹配，然后其他需要匹配的就是w1的前i个以及w2的前j － 1个。 f(i, j ) = f (i, j - 1) + 1
2, del, 删除后，需要匹配的就是w1前i－1位跟w2的前j位。 f(i, j) = f(i - 1, j) + 1
3，replace, 换了这个之后，相当于直接匹配这两个位置的字母，需要匹配剩下，i - 1,j -1个。f(i, j) = f (i - 1,j - 1) + 1
总的方程就是：if(w1(i) == w2(j)) ==> f(i-1,j -1)
else ==> min(insert,del, replace)
初始化：值得注意的是这里，f(i, 0) = i 以及 f(0, i) = i 意思就是w1或者w2为空时候的情况。
结果：f（w1.len，w2.len）
代码：
public class Solution { /** * @param word1 & word2: Two string. * @return: The minimum number of steps. */ // insert f(i, j -1) + 1 // del f(i -1, j) + 1 // replace f(i - 1, j - 1) + 1 // word1(i) = word2(j) f(i - 1, j - 1); // f(0, 0) = 0; public int minDistance(String word1, String word2) { if (word1.length() == 0 || word1 == null){ return word2.length(); } if (word2.length() == 0 || word2 == null){ return word1.length(); } int[][] result = new int[word1.length() + 1][word2.length() + 1]; for (int i = 0; i <= word1.length(); i++){ result[i][0] = i; } for (int i = 0; i<= word2.length(); i ++){ result[0][i] = i; } for (int i = 1; i <= word1.length(); i++){ for (int j = 1; j <= word2.length(); j++){ if (word1.charAt(i - 1) == word2.charAt(j - 1)){ result[i][j] = result[i - 1][j - 1]; } else { result[i][j] = Math.min(result[i - 1][j - 1], Math.min(result[i - 1][j], result[i][j - 1])) + 1; } } } return result[word1.length()][word2.length()]; } }
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原文地址：http://www.cnblogs.com/tritritri/p/4955100.html
