题目
给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。
你可以对一个单词进行如下三种操作：
插入一个字符
删除一个字符
替换一个字符示例 1:
输入: word1 = "horse", word2 = "ros"
输出: 3
解释:
horse -> rorse (将 ‘h‘ 替换为 ‘r‘)
rorse -> rose (删除 ‘r‘)
rose -> ros (删除 ‘e‘)示例 2:
输入: word1 = "intention", word2 = "execution"
输出: 5
解释:
intention -> inention (删除 ‘t‘)
inention -> enention (将 ‘i‘ 替换为 ‘e‘)
enention -> exention (将 ‘n‘ 替换为 ‘x‘)
exention -> exection (将 ‘n‘ 替换为 ‘c‘)
exection -> execution (插入 ‘u‘)题解
这个题目拿到题目基本就能想到DP，因为感觉和我们之前的爬楼梯啥的比较相似。这个题目比较为hard主要是，状态转换比较复杂。
定义: dpi ， word1这个字符串的前i个 -> word1这个字符串的前j 个字符，所需要的最小的步数
那么有以下几种情况
word1[i] == word2[j]
不需要做变化，那么 dpi = dp[i-1,j-1]
word1[i] != word2[j]
我们就需要动用上面那三种操作了：
- add = dp[i, j-1], 代表插入一个字符
- delete = dp[i-1, j]，代表删除一个字符
- replace = dp[i-1, j-1]，代表替换一个字符
- dpi = 1 + min(add, delete, replace)
时间复杂度 o(m * n)
java
class Solution {
public int minDistance(String word1, String word2) {
int m = word1.length();
int n = word2.length();
int[][] dp = new int[m + 1][n + 1];
// base
for (int i = 0; i <= m; i++) {
dp[i][0] = i;
}
for (int j = 0; j <= n; j++) {
dp[0][j] = j;
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (word1.charAt(i) == word2.charAt(j)) {
dp[i + 1][j + 1] = dp[i][j];
} else {
int add = dp[i][j + 1];
int delete = dp[i + 1][j];
int rep = dp[i][j];
dp[i + 1][j + 1] = Math.min(Math.min(add, delete), rep) + 1;
}
}
}
return dp[m][n];
}
}python
class Solution:
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
m = len(word1)
n = len(word2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(m + 1):
dp[i][0] = i
for i in range(n + 1):
dp[0][i] = i
for i in range(m):
for j in range(n):
if word1[i] == word2[j]:
dp[i + 1][j + 1] = dp[i][j]
else:
add = dp[i][j + 1]
delete = dp[i + 1][j]
replace = dp[i][j]
dp[i + 1][j + 1] = min(add, delete, replace) + 1
return dp[m][n]
