LeetCode - Edit Distance

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

思路：

动态规划，dp(i,j)指(i-1)和(j-1)位置需要多少次变换才能相等，若word1[i-1]==word2[j-1]，那么dp(i,j)=dp(i-1,j-1)，否则的话，则是从dp(i-1,j-1)，dp(i,j-1)，dp(i-1,j)中选择一个最小值，然后再加1.

package dp;

public class EditDistance {

public int minDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();
int[][] dp = new int[m + 1][n + 1];
for (int i = 0 ;i <= m ; ++i)
dp[i][0] = i;
for (int i = 0; i <= n; ++i)
dp[0][i] = i;
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = 1 + Math.min(Math.min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]);
}
}
}

return dp[m]
;
}

public static void main(String[] args) {
// TODO Auto-generated method stub
EditDistance e = new EditDistance();
System.out.println(e.minDistance("a", "b"));
System.out.println(e.minDistance("abc", "bc"));
System.out.println(e.minDistance("abc", "def"));
System.out.println(e.minDistance("abcdefg", "bcdefgh"));
System.out.println(e.minDistance("abc", "cba"));

}

}