Problem Description:
Given two strings A and B, use the least operation to convert a string A string B. Wherein the string comprises:
(1) deleting a character (the Insert Character A)
(2) is inserted into a character (the Delete Character A)
(. 3) to modify a character (Replace a character)
A string converted into strings B operands with minimum character string called edit distance A to B, also known as Levenshtein distance, in 1965, Russian mathematician Vladimir Levenshtein proposed.
问题分析：动态规划思想
（1）、dp[i][j]表示将字符串 A[0: i-1] 转变为 B[0: j-1] 的最小步骤数。
（2）、边界情况：
当 i = 0 , 即 A 串为空时，那么转变为 B 串就是不断添加字符，dp[0][j] = j。
当 j = 0，即 B 串为空时，那么转变为 B 串就是不断删除字符，dp[i][0] = i。
（3）、对应三种字符操作方式：
插入操作：dp[i][j - 1] + 1 相当于为 B 串的最后插入了 A 串的最后一个字符；
删除操作：dp[i - 1][j] + 1 相当于将 B 串的最后字符删除 ;
替换操作：dp[i - 1][j - 1] +（A[i - 1] != B[j - 1]）相当于通过将 B 串的最后一个字符替换为 A 串的最后一个字符。
（4）所以dp方程式为：
1<=i<=len(A)， 1<=j<=len(B)
dp[i][j] = dp[i-1][j-1], A[i-1] == B[j-1]
dp[i][j] = min(dp[i-1][j-1]+1, dp[i][j-1]+1, dp[i-1][j]+1), A[i-1] != B[j-1]
class Solution:
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
m, n = len(word1), len(word2)
if m == 0:return n
if n == 0:return m
dp = [[0]*(n+1) for _ in range(m+1)] # 初始化dp和边界
for i in range(1, m+1): dp[i][0] = i
for j in range(1, n+1): dp[0][j] = j
for i in range(1, m+1): # 计算dp
for j in range(1, n+1):
if word1[i - 1] == word2[j - 1]:
dp[i][j] = dp[i - 1][j - 1]
else:
dp[i][j] = min(dp[i - 1][j - 1] + 1, dp[i][j - 1] + 1, dp[i - 1][j] + 1)
return dp[m][n]
if __name__ == '__main__':
solu = Solution()
word1, word2 = 'horse', 'ros'
print(solu.minDistance(word1, word2))