编辑距离 困难🌟🌟🌟🌟
问题描述
原文链接：72. 编辑距离
给你两个单词 word1 和 word2， 请返回将 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')
提示：
0 <= word1.length, word2.length <= 500word1和word2由小写英文字母组成
代码实现
Java
class Solution {
/*
1.dp[i][j]：表示长度为 i 的word1转为长度为 j 的word2所需要的最少操作次数为dp[i][j]。
2. 关系式
word1 = hor
word2 = ho
dp[i][j] =
(1). w1[i] == w2[j]=>dp[i][j] = dp[i-1][j-1]。
(2).如果 w1[i] != w2[j]
a. 插入：dp[i][j] = dp[i][j-1] + 1。
b.删除 ：dp[i][j] = dp[i-1][j] + 1
c.替换： dp[i][j] = dp[i-1][j-1] + 1.
dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1
3、初始值
dp[0][j] = j
dp[i][0] = i
*/
public int minDistance(String word1, String word2) {
int n1 = word1.length();
int n2 = word2.length();
int[][] dp = new int[n1+1][n2+1];
// 初始化
for(int i = 0; i <= n1; i++){
dp[i][0] = i;
}
for(int j = 0; j <= n2; j++){
dp[0][j] = j;
}
//
for(int i = 1; i <= n1; i++){
for(int j = 1; j <= n2; j++){
if(word1.charAt(i-1) == word2.charAt(j - 1)){
dp[i][j] = dp[i-1][j-1];
}else{
dp[i][j] = Math.min(Math.min(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1;
}
}
}
return dp[n1][n2];
}
}
Python
class Solution(object):
def minDistance(self, word1, word2):
"""
:type word1: str
:type word2: str
:rtype: int
"""
n1 = len(word1)
n2 = len(word2)
dp = [[0] * (n2+1) for _ in range(n1+1)]
# 初始化
for i in range(n1+1):
dp[i][0] = i
for j in range(n2+1):
dp[0][j] = j
# DP迭代
for i in range(1, n1+1):
for j in range(1, n2+1):
if word1[i-1] == word2[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1
return dp[n1][n2]
C++
class Solution {
public:
int minDistance(string word1, string word2) {
int n1 = word1.length();
int n2 = word2.length();
vector<vector<int>> dp(n1+1, vector<int>(n2+1));
// 初始化
for(int i = 0; i <= n1; i++){
dp[i][0] = i;
}
for(int j = 0; j <= n2; j++){
dp[0][j] = j;
}
// DP迭代
for(int i = 1; i <= n1; i++){
for(int j = 1; j <= n2; j++){
if(word1[i-1] == word2[j-1]){
dp[i][j] = dp[i-1][j-1];
}else{
dp[i][j] = min(min(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1;
}
}
}
return dp[n1][n2];
}
};
Go
func minDistance(word1 string, word2 string) int {
n1 := len(word1)
n2 := len(word2)
dp := make([][]int, n1+1)
for i := range dp {
dp[i] = make([]int, n2+1)
}
// 初始化
for i := 0; i <= n1; i++ {
dp[i][0] = i
}
for j := 0; j <= n2; j++ {
dp[0][j] = j
}
// DP迭代
for i := 1; i <= n1; i++ {
for j := 1; j <= n2; j++ {
if word1[i-1] == word2[j-1] {
dp[i][j] = dp[i-1][j-1]
} else {
dp[i][j] = min(min(dp[i][j-1], dp[i-1][j]), dp[i-1][j-1]) + 1
}
}
}
return dp[n1][n2]
}
func min(a, b int) int {
if a < b {
return a
}
return b
}