[leetcode DP]72. Edit Distance

计算最少用多少不把word1变为word2，

思路：建立一个dp表，行为word1的长度，宽为word2的长度

1.边界条件，dp[i][0] = i,dp[0][j]=j

2.最优子问题，考虑已经知道word1[0:i-1]转变为word2[0:j-1]的次数，只需要考虑word1[i]和word[j]的情况

3.子问题重叠，word1[i]和word2[j]是否相等，每种情况下怎样有最少步骤

 1 class Solution(object):
 2     def minDistance(self, word1, word2):
 3         d1,d2 = len(word1),len(word2)
 4         dp = [[0 for j in range(d2+1)] for i in range(d1+1)]
 5         for i in range(d1+1):
 6             dp[i][0] = i
 7         for j in range(d2+1):
 8             dp[0][j] = j
 9         for i in range(1,d1+1):
10             for j in range(1,d2+1):
11                 if word1[i-1] == word2[j-1]:
12                     dp[i][j] = dp[i-1][j-1]
13                 else:
14                     dp[i][j] = min(dp[i][j-1],dp[i-1][j-1],dp[i-1][j])+1
15         return dp[d1][d2]
16

 

posted @ 2017-03-14 10:30  wilderness  阅读(165)  评论(0)  编辑  收藏  举报