Edit Distance

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。

public class Solution {
    public int minDistance(String word1, String word2) {
    int len1 = word1.length();
    int len2 = word2.length();
 
    // len1+1, len2+1, because finally return dp[len1][len2]
    int[][] dp = new int[len1 + 1][len2 + 1];
 
    for (int i = 0; i <= len1; i++) 
        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 delete = dp[i-1][j] + 1;
                int insert = dp[i][j-1] + 1;
 
                int min = Math.min(replace, insert);
                min = Math.min(min,delete);
                dp[i][j] = min;
            }
        }
    }
 
    return dp[len1][len2];
}
}