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Edit Distance

时间:2014-06-30 00:46:38      阅读:262      评论:0      收藏:0      [点我收藏+]

标签:java   leetcode   string   dp   

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

方法

经典的编辑距离问题。
此处讲解清晰:http://www.cnblogs.com/biyeymyhjob/archive/2012/09/28/2707343.html
    public int minDistance(String word1, String word2) {
        if (word1 == null || word2 == null) {
            return 0;
        }
        int len1 = word1.length();
        int len2 = word2.length();
        if (len1 == 0 || len2 == 0) {
            return Math.max(len1, len2);
        }
        
        int[][] status = new int[len1 + 1][len2 + 1];
        for (int i = 0; i < len1 + 1; i++) {
            for (int j = 0; j < len2 + 1; j++) {
                if (i == 0) {
                    status[i][j] = j;
                } else if (j == 0) {
                    status[i][j] = i;
                } else {
                    int tempMin = Math.min(status[i - 1][j], status[i][j - 1]) + 1;
                    if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                        status[i][j] = Math.min(tempMin, status[i - 1][j - 1]);
                    } else {
                        status[i][j] = Math.min(tempMin, status[i - 1][j - 1] + 1);
                     }
                }
            }
        }
        return status[len1][len2];
    }


Edit Distance,布布扣,bubuko.com

Edit Distance

标签:java   leetcode   string   dp   

原文地址:http://blog.csdn.net/u010378705/article/details/35566983

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