Edit Distance

1.题目

给出两个单词word1和word2，计算出将word1 转换为word2的最少操作次数。

你总共三种操作方法：

• 插入一个字符
• 删除一个字符
• 替换一个字符

样例

给出 work1="mart" 和 work2="karma"

返回 3

2.算法

如果我们用 i 表示当前字符串 A 的下标，j 表示当前字符串 B 的下标。 如果我们用d[i, j] 来表示A[1, ... , i] B[1, ... , j] 之间的最少编辑操作数。那么我们会有以下发现：

1. d[0, j] = j;

2. d[i, 0] = i;

3. d[i, j] = d[i-1, j - 1] if A[i] == B[j]

4. d[i, j] = min(d[i-1, j - 1], d[i, j - 1], d[i-1, j]) + 1  if A[i] != B[j]

    public static int minDistance(String word1, String word2) {          if (word1.length() == 0) return word2.length();          if (word2.length() == 0) return word1.length();                    int[][] distance = new int[word1.length() + 1][word2.length() + 1];                    for (int i = 0; i <= word1.length(); i++) {              distance[i][0] = i;          }                    for (int i = 0; i <= word2.length(); i++) {              distance[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)) {                      distance[i][j] = distance[i-1][j-1];                  } else {                    distance[i][j] = Math.min(distance[i-1][j], Math.min(distance[i][j-1], distance[i-1][j-1])) + 1;                 }  //替换和插入            }          }          return distance[word1.length()][word2.length()];      }  

    public static int minDistance(String word1, String word2)     {          if (word1.length() == 0) return word2.length();          if (word2.length() == 0) return word1.length();          String minWord = word1.length()>word2.length()?word2:word1;          String maxWord = word1.length()>word2.length()?word1:word2;          int[] res = new int[minWord.length() + 1];        for (int i = 0; i <= minWord.length(); i++)        {        res[i] = i;  //当另一个字符串为空时，需改变i ＋　１次        }        for (int i = 0; i < maxWord.length(); i++)//res[i][j] = min(res[i-1][j], res[i][j-1], res[i-1][j-1])+1。        {            int[] newRes = new int[minWord.length() + 1];　//对每个i建立一个数组j            newRes[0] = i + 1;            for (int j = 0; j < minWord.length(); j++)            {            if (minWord.charAt(j) == maxWord.charAt(i))  //res[i][j]=res[i-1][j-1]            {            newRes[j + 1] = res[j];             }            else             {   //res[i][j] = min(res[i-1][j], res[i][j-1], res[i-1][j-1])+1。                    newRes[j+1] = Math.min(res[j],Math.min(res[j+1],newRes[j]))+1;             }            }            res = newRes;        }        return res[minWord.length()];    }