标签:style blog http color os art
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
[leetcode] Edit Distance,布布扣,bubuko.com
标签:style blog http color os art
原文地址:http://www.cnblogs.com/jdflyfly/p/3812776.html