[leetcode]Edit Distance

动态规划，此题思考许久，出错不少。动态规划的转移方程倒是想出来了，转移方程的讨论可见：http://blog.unieagle.net/2012/09/19/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Aedit-distance%EF%BC%8C%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B9%8B%E9%97%B4%E7%9A%84%E7%BC%96%E8%BE%91%E8%B7%9D%E7%A6%BB%EF%BC%8C%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92/

但是对于i==0和j==0时的计算，以至于a和aa，以及a和ba这样的例子都计算错过。

后来看到一个用长度而非坐标来作为数组索引的，就更简单了：http://www.cnblogs.com/remlostime/archive/2012/11/06/2757668.html

public class Solution {
    public int minDistance(String word1, String word2) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int len1 = word1.length();
        int len2 = word2.length();
        if (len1 == 0) return len2;
        if (len2 == 0) return len1;
        int m[][] = new int[len1][len2];
        
        m[0][0] = word1.charAt(0) == word2.charAt(0) ? 0 : 1;
        for (int j = 1; j < len2; j++) {
            if (word1.charAt(0) == word2.charAt(j)) {
                m[0][j] = j;
            }
            else {
                m[0][j] = m[0][j-1]+1;
            }
        }
        
        for (int i = 1; i < len1; i++) {
            if (word1.charAt(i) == word2.charAt(0)) {
                m[i][0] = i;
            }
            else {
                m[i][0] = m[i-1][0]+1;
            }
        }
        
        for (int i = 1; i < len1; i++) {
            for (int j = 1; j < len2; j++) {
                int x1 = m[i-1][j] + 1;
                int x2 = m[i][j-1] + 1;
                int x3 =  m[i-1][j-1];
                if (word1.charAt(i) != word2.charAt(j)) x3++;
                m[i][j] = Math.min(x1, Math.min(x2, x3));               
            }
        }
        return m[len1-1][len2-1];
    }
}

　　

转载于:https://www.cnblogs.com/lautsie/p/3251311.html