【LeetCode with Python】 Edit Distance

博客域名：http://www.xnerv.wang
原题页面：https://oj.leetcode.com/problems/edit-distance/
题目类型：动态规划
难度评价：★★★
本文地址：http://blog.csdn.net/nerv3x3/article/details/37334113

Given two words word1 and word2, find the minimum number of steps required to convertword1 toword2. (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

一道传统意义上的经典的动态规划题目。注意动态规划十有八九可以优化空间复杂度，不是真的要开辟一个矩阵出来。

class Solution:    # @return an integer    def minDistance(self, word1, word2):        if "" == word1 and "" == word2:            return 0        elif "" == word1:            return len(word2)        elif "" == word2:            return len(word1)        len1 = len(word1)        len2 = len(word2)        arr1 = [0 for y in range(0, len1 + 1)]        arr2 = [0 for y in range(0, len1 + 1)]        arr1[0] = 0        for y in range(1, len1 + 1):            arr1[y] = y        for x in range(1, len2 + 1):            arr2[0] = x            for y in range(1, len1 + 1):                arr2[y] = min(arr1[y - 1] + (0 if (word1[y - 1] == word2[x - 1]) else 1), arr1[y] + 1, arr2[y - 1] + 1)            tmp = arr1            arr1 = arr2            arr2 = tmp            for y in range(0, len1 + 1):                arr2[y] = 0        return arr1[len1]