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LeetCode - Edit Distance

时间:2015-08-27 23:17:06      阅读:420      评论:0      收藏:0      [点我收藏+]

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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

一个非常经典的动态规划问题,状态转移方程:

                                         f(i-1, j-1)                                                        word1[i] == word2[j]

                        f(i, j)   = 

                                         Min{ f(i-1, j), f(i, j-1), f(i-1, j-1} }  + 1          word1[i] != word2[j]

如上所示,f(i, j)表示word1.substring(i)与word2.substring(j)直接的距离

1)显然,如果word1[i] == word2[j],转化为递归解,即f(i-1, j-1)

2)如果word1[i] == word2[j],则可以对两个字符的最后一个字符串进行三种操作,对与字符word1[i]可以进行如下三种操作:

      i) delete,即把此字符删除,则问题转化为f(i-1, j)

      ii) replace,即把此字符替换为word2[j],则问题转化为f(i-1, j-1)

     iii) Insert,即在此处增加一个字符word2[j], 则问题转化为f(i, j-1)

根据上述状态转移方程,很容易写出代码

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

又是一个比较经典的动态规划问题。

版权声明:本文为博主原创文章,未经博主允许不得转载。

LeetCode - Edit Distance

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原文地址:http://blog.csdn.net/my_jobs/article/details/48032665

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