EditDistance.java

package leetcode;

/**
 * Project Name : Leetcode
 * Package Name : leetcode
 * File Name : EditDistance
 * Creator : Edward
 * Date : Dec, 2017
 * Description : 72. Edit Distance
 */
public class EditDistance {
    /**
     * 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

     abcd -> aef

     dp[i][j]表示的是，从字符串1的i的位置转换到字符串2的j的位置，所需要的最少步数。


     1,字符串中的字符相等: dp[i][j] = dp[i - 1][j - 1]
     2,字符串中的字符不等:
         insert: dp[i][j] = dp[i][j - 1] + 1;
         replace: dp[i][j] = dp[i - 1][j - 1] + 1;
         delete: dp[i][j] = dp[i - 1][j] + 1;

           a  b  c  d
        0  1  2  3  4
     a  1  0  1  2  3
     e  2  1  1  2  3
     f  3  2  2  2  3

     time : O(m * n)
     space : O(m * n)

     * @param word1
     * @param word2
     * @return
     */
    public int minDistance(String word1, String word2) {
        int len1 = word1.length();
        int len2 = word2.length();

        int[][] dp = new int[len1 + 1][len2 + 1];
        for (int i = 0; i <= len1; i++) {
            dp[i][0] = i;
        }
        for (int i = 0; i <= len2; i++) {
            dp[0][i] = i;
        }
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i -1][j - 1]);
                }
            }
        }
        return dp[len1][len2];
    }
}