动态规划的问题.建立一个二维数组ans，其中ans[i][j]代表word1[i：]要变成word2[j：]至少需要多少步。那么如果word1[i] == word2[j]，则ans[i][j] = ans[i+1][j+1]，否者，ans[i][j]= min（ans[i +1][j]，ans[ans[i][j+1],ans[i+1][j+1]） + 1。
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
m,n = len(word1),len(word2)
ans = [[0 for i in range(n + 1)] for j in range(m + 1)]
for i in range(m + 1):
ans[i][n] = m - i
for i in range(n + 1):
ans[m][i] = n - i
m -= 1;n -= 1
while m >= 0:
t = n
while t >= 0:
if word1[m] == word2[t]:
ans[m][t] = ans[m + 1][t + 1]
else:
ans[m][t] = min(ans[m][t+1],ans[m+1][t],ans[m+1][t+1]) + 1
t -= 1
m -= 1
return ans[0][0]
