LeetCode-Edit_Distance
比较特殊的动态规划过程，原题链接：https://leetcode.com/problems/edit-distance/
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
# _*_ coding: utf-8 _*_
class Solution(object):
def minDistance1(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"A B G C _ F"
"_ B _ D E _"
"""
n = len(word1)
m = len(word2)
if not m or not n:
return m or n
dp = [[0] * (n+1) for i in xrange(m+1)]
for i in xrange(m+1):
dp[i][0] = i
for j in xrange(n+1):
dp[0][j] = j
for i in xrange(m):
for j in xrange(n):
if word1[j] != word2[i]:
temp = 1
else:
temp = 0
dp[i+1][j+1] = min(dp[i][j+1]+1, dp[i+1][j]+1, dp[i][j]+temp)
return dp[m][n]
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
减少了空间复杂度，时间复杂度依然为O(m*n)
"""
n = len(word1)
m = len(word2)
if not m or not n:
return m or n
dp1, dp2 = [0] * (m+1), [i for i in xrange(m+1)]
for i in xrange(n):
dp1, dp2 = dp2, [i+1] * (m+1)
for j in xrange(m):
if word1[i] != word2[j]:
diff = 1
else:
diff = 0
dp2[j+1] = min(dp1[j]+diff, dp1[j+1]+1, dp2[j]+1)
return dp2[m]
虽然这两种用法我都已经解出，但是回忆起又一次在线笔试题，计算编辑的成本我就细思极恐，比如插入一个字符消耗1个能量，删除一个字符消耗2个能量，替换一个字符消耗3个能量，给定两个字符串求最少的编辑能耗…待我想出来了再给出解题代码，知道解法的大神也请联系我，已跪！