72. 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

链接：  http://leetcode.com/problems/edit-distance/

题解：

Dynamic Programming动态规划的经典问题，一定要好好继续研究一下。 详解请看下面的reference。 还可以使用滚动数组继续优化空间为O(n)或者O(m)。最近在忙于房子装修，都没有时间刷题和准备面试，下一遍要补上。

下周一onsite BB，裸面，希望有好运气吧！

Time Complexity - O(mn)， Space Complexity - O(mn)。

public class Solution {
    public int minDistance(String word1, String word2) {
        if(word1 == null || word2 == null)
            return 0;
        int word1Len = word1.length(), word2Len = word2.length();
        int[][] dp = new int[word1Len + 1][word2Len + 1];
        
        for(int i = 0; i < word1Len + 1; i++)       //word1 as row
            dp[i][0] = i;
        
        for(int j = 1; j < word2Len + 1; j++)       //word2 as column
            dp[0][j] = j;
            
        for(int i = 1; i < word1Len + 1; i++) {
            for(int j = 1; j < word2Len + 1; 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(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j]));
            }
        }
        
        return dp[word1Len][word2Len];
    }
}

Update:

主要使用DP，假设以word1为列，word2为行，初始化的时候设定distance[0][i]以及distance[j][0] - 当对方字符串为空时需要多少步骤。则转移方程为，当前字符相同时，distance[i][j] = distance[i - 1][j - 1]， 否则这时insert， replace，delete权重都为1， 方程为1 + 三种改变的最小值， 既Math.min(distance[i - 1][j - 1]， Math.min(distance[i - 1][j]， distance[i][j - 1]))。 其中distance[i - 1][j - 1]为replace， distance[i - 1][j]是word1删除一个字符， distance[i][j - 1]是word2删除一个字符。

public class Solution {
    public int minDistance(String word1, String word2) {
        if(word1 == null || word2 == null)
            return 0;
        int word1Len = word1.length(), word2Len = word2.length();
        int[][] distance = new int[word1Len + 1][word2Len + 1];
        
        for(int i = 1; i < word1Len + 1; i++)
            distance[i][0] = i;
            
        for(int j = 1; j < word2Len + 1; j++)
            distance[0][j] = j;
        
        for(int i = 1; i < word1Len + 1; i++) {
            for(int j = 1; j < word2Len + 1; j++)
                if(word1.charAt(i - 1) == word2.charAt(j - 1))
                    distance[i][j] = distance[i - 1][j - 1];
                else
                    distance[i][j] = 1 + Math.min(distance[i - 1][j - 1], Math.min(distance[i - 1][j], distance[i][j - 1]));
        }
            
        return distance[word1Len][word2Len];
    }
}

二刷

思路仍然不是特别清晰。我们尝试分为以下几个步骤:

1. 这道题目应该使用dp。
2. 要解决的是如何定义dp,  如何设置初始化状态，以及转移方程是什么。
3. 首先我们考虑边界条件，当有一个string为空的时候我们返回0。
4. 接下来创建一个dp矩阵dist，假如word1的长度为word1Len，word2的长度为word2Len，那么这个矩阵的长度就为[word1Len + 1， word2Len + 1]。
5. 我们初始化第一行和第一列，dist[i][0] = i， dist[0][j] = j，  都是负责处理其中一个word为空这种情况。
6. 接下来，我们定义dist[i][j]为 word1(0, i) 到word2(0,j) 这两个单词的min Edit distance。那么我们有以下的公式:
   1. 假如word1.charAt(i) == word2.charAt(j)，那么dist[i][j] = 0
   2. 否则dist[i][j] = 1 + min (dist[i - 1][j - 1], min(dist[i - 1][j], dist[i][j - 1]))。
      1. 这里假如使用dist[i - 1][j - 1]，意思是replace
      2. 假如使用dist[i - 1][j]，那么是word1比word2少1个字符。 对word1来说是add
      3. 假如使用dist[i][j - 1]，那么是word2比word1多一个字符。对word1来说是delete
7. 最后返回结果dist[word1Len][word2Len]
8. 这里其实也可以简化为滚动数组，达到Space Complexity - O(n)的结果，留给三刷了。

Java:

Time Complexity - O(mn)， Space Complexity - O(mn)。

public class Solution {
    public int minDistance(String word1, String word2) {
        if (word1 == null || word2 == null) {
            return 0;
        }
        int word1Len = word1.length(), word2Len = word2.length();
        int[][] dist = new int[word1Len + 1][word2Len + 1];
        for (int i = 1; i <= word1Len; i++) {
            dist[i][0] = i;
        }
        for (int j = 1; j <= word2Len; j++) {
            dist[0][j] = j;
        }
        
        for (int i = 1; i <= word1Len; i++) {
            for (int j = 1; j <= word2Len; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dist[i][j] = dist[i - 1][j - 1];
                } else {
                    dist[i][j] = Math.min(dist[i - 1][j - 1], Math.min(dist[i - 1][j], dist[i][j - 1])) + 1;    
                }
            }
        }
        
        return dist[word1Len][word2Len];
    }
}

三刷:

还是dp。当两字符相等时，取左上的值。 否则表示有一个edit distance，我们取左上，上和左三个值里最小的一个，+ 1，然后继续计算。

Java:

public class Solution {
    public int minDistance(String word1, String word2) {
        if (word1 == null || word2 == null) return Integer.MAX_VALUE;
        int m = word1.length(), n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) dp[i][0] = i;
        for (int j = 1; j <= n; j++) dp[0][j] = j;
        
        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(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1]));
            }
        }
        return dp[m][n];
    }
}

一维DP:

跟Maximal Square一样，也是使用一个topLeft来代表左上方的元素。

public class Solution {
    public int minDistance(String word1, String word2) {
        if (word1 == null || word2 == null) return Integer.MAX_VALUE;
        int m = word1.length(), n = word2.length();
        if (m == 0) return n;
        else if (n == 0) return m;
        
        int[] dp = new int[n + 1];
        for (int j = 1; j <= n; j++) dp[j] = j;
        int topLeft = 0;
        
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                int tmp = dp[j];
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) dp[j] =  topLeft;
                else dp[j] = 1 + Math.min(topLeft, Math.min(dp[j], dp[j - 1]));
                topLeft = tmp;
            }
            dp[0] = i;
            topLeft = i;
        }
        return dp[n];
    }
}

Reference：

https://leetcode.com/discuss/10426/my-o-mn-time-and-o-n-space-solution-using-dp-with-explanation

http://www.cnblogs.com/springfor/p/3896167.html

https://leetcode.com/discuss/17997/my-accepted-java-solution

https://leetcode.com/discuss/20945/standard-dp-solution

https://leetcode.com/discuss/5138/good-pdf-on-edit-distance-problem-may-be-helpful

https://leetcode.com/discuss/43398/20ms-detailed-explained-c-solutions-o-n-space

http://web.stanford.edu/class/cs124/lec/med.pdf

https://en.wikipedia.org/wiki/Edit_distance

https://leetcode.com/discuss/64063/ac-python-212-ms-dp-solution-o-mn-time-o-n-space

https://leetcode.com/discuss/43398/20ms-detailed-explained-c-solutions-o-n-space