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[leetcode] Edit Distance
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
https://oj.leetcode.com/problems/edit-distance/
分析：编辑距离，算法导论课后题，编程之美上好像都有。概念的话，看 这里
思路1：dfs，复杂度指数级别，目测要超时。
思路2: 二维的DP，详见参考1，分析的很好。
public class Solution { public int minDistance(String word1, String word2) { int length1 = word1.length(); int length2 = word2.length(); if (length1 == 0 || length2 == 0) { return length1 == 0 ? length2 : length1; } int[][] distance = new int[length1 + 1][length2 + 1]; distance[0][0] = 0; for (int i = 1; i <= length1; i++) { distance[i][0] = i; } for (int i = 1; i <= length2; i++) { distance[0][i] = i; } for (int i = 1; i <= length1; i++) { for (int j = 1; j <= length2; 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 - 1], Math.min(distance[i][j - 1], distance[i - 1][j])) + 1; } } } return distance[length1][length2]; } public static void main(String[] args) { System.out.println(new Solution().minDistance("eeba", "abca")); }}
参考：
http://huntfor.iteye.com/blog/2077940
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/
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