题目
Given two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.
In one step, you can delete exactly one character in either string.
Example 1:
Input: word1 = "sea", word2 = "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".
Example 2:
Input: word1 = "leetcode", word2 = "etco"
Output: 4
Constraints:
1 <= word1.length, word2.length <= 500word1andword2consist of only lowercase English letters.
思路
为word1和word2分别设两个指针i、j进行遍历，总共分为如下几种情况：
- i遍历到底且j遍历到底：无需删除任何字符。
- i遍历到底或j遍历到底：有一方已经遍历到底，需删除一方剩余的字符。
- i和j所标识的字符相等：当前字符相等，无需删除，i和j均前进一步。
- i和j所标识的字符不相等：分为删除i字符或删除j字符两种情况，随后删除的一方前进一步。
代码
python版本：
# dfs，超时
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
def dfs(i, j):
if i == len(word1) and j == len(word2):
return 0
if i == len(word1) or j == len(word2):
return len(word1)+len(word2)-i-j
if word1[i] == word2[j]:
return dfs(i+1, j+1)
return min(dfs(i+1, j), dfs(i, j+1))+1
return dfs(0, 0)
# 优化dfs，使用辅助数组记录重复调用 1255 ms
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
helper = [[-1]*len(word2) for _ in range(len(word1))]
def dfs(i, j):
print(i, j)
if i == len(word1) and j == len(word2):
return 0
if i == len(word1) or j == len(word2):
return len(word1)+len(word2)-i-j
if helper[i][j] != -1:
return helper[i][j]
if word1[i] == word2[j]:
helper[i][j] = dfs(i+1, j+1)
return helper[i][j]
helper[i][j] = min(dfs(i+1, j), dfs(i, j+1))+1
return helper[i][j]
res = dfs(0, 0)
return res
# 动态规划 320 ms
class Solution:
def minDistance(self, word1: str, word2: str) -> int:
dp = [[math.inf]*(len(word2)+1) for _ in range(len(word1)+1)]
for i in range(len(dp)):
for j in range(len(dp[i])):
if i == 0 or j == 0:
dp[i][j] = i or j
elif word1[i-1] == word2[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = min(dp[i-1][j], dp[i][j-1])+1
return dp[-1][-1]
