@@ -51,27 +51,134 @@ exection -> execution (插入 'u')
5151 <li><code>word1</code> 和 <code>word2</code> 由小写英文字母组成</li>
5252</ul >
5353
54-
5554## 解法
5655
5756<!-- 这里可写通用的实现逻辑 -->
5857
58+ 动态规划。
59+
60+ 设 ` dp[i][j] ` 表示将 word1 前 i 个字符组成的字符串 ` word1[0...i-1] ` 转换成 word2 前 j 个字符组成的字符串 ` word2[0...j-1] ` 的最小操作次数。
61+
62+ 初始化 ` dp[i][0] = i ` （` i∈[0, word1.length] ` ），` dp[0][j] = j ` （` j∈[0, word2.length] ` ）。
63+
64+ i, j 分别从 1 开始遍历，判断 ` word1[i - 1] ` 与 ` word2[j - 1] ` 是否相等：
65+
66+ - 若 ` word1[i - 1] == word2[j - 1] ` ，` dp[i][j] = dp[i - 1][j - 1] ` 。
67+ - 若 ` word1[i - 1] != word2[j - 1] ` ，` dp[i][j] = min(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1]) + 1 ` 。其中 ` dp[i - 1][j] + 1 ` 对应插入操作，` dp[i][j - 1] + 1 ` 对应删除操作，` dp[i - 1][j - 1] + 1 ` 对应替换操作。取三者的最小值即可。
68+
5969<!-- tabs:start -->
6070
6171### ** Python3**
6272
6373<!-- 这里可写当前语言的特殊实现逻辑 -->
6474
6575``` python
66-
76+ class Solution :
77+ def minDistance (self word1 : str , word2 : str ) -> int :
78+ m, n = len (word1), len (word2)
79+ dp = [[0 ] * (n + 1 ) for _ in range (m + 1 )]
80+ for i in range (m + 1 ):
81+ dp[i][0 ] = i
82+ for j in range (n + 1 ):
83+ dp[0 ][j] = j
84+ for i in range (1 , m + 1 ):
85+ for j in range (1 , n + 1 ):
86+ if word1[i - 1 ] == word2[j - 1 ]:
87+ dp[i][j] = dp[i - 1 ][j - 1 ]
88+ else :
89+ dp[i][j] = min (dp[i][j - 1 ], dp[i - 1 ]
90+ [j], dp[i - 1 ][j - 1 ]) + 1
91+ return dp[- 1 ][- 1 ]
6792```
6893
6994### ** Java**
7095
7196<!-- 这里可写当前语言的特殊实现逻辑 -->
7297
7398``` java
99+ class Solution {
100+ public int minDistance (String word1 , String word2 ) {
101+ int m = word1. length(), n = word2. length();
102+ int [][] dp = new int [m + 1 ][n + 1 ];
103+ for (int i = 0 ; i <= m; ++ i) {
104+ dp[i][0 ] = i;
105+ }
106+ for (int j = 0 ; j <= n; ++ j) {
107+ dp[0 ][j] = j;
108+ }
109+ for (int i = 1 ; i <= m; ++ i) {
110+ for (int j = 1 ; j <= n; ++ j) {
111+ if (word1. charAt(i - 1 ) == word2. charAt(j - 1 )) {
112+ dp[i][j] = dp[i - 1 ][j - 1 ];
113+ } else {
114+ dp[i][j] = Math . min(Math . min(dp[i][j - 1 ], dp[i - 1 ][j]), dp[i - 1 ][j - 1 ]) + 1 ;
115+ }
116+ }
117+ }
118+ return dp[m][n];
119+ }
120+ }
121+ ```
122+
123+ ### ** C++**
124+
125+ ``` cpp
126+ class Solution {
127+ public:
128+ int minDistance(string word1, string word2) {
129+ int m = word1.size(), n = word2.size();
130+ vector<vector<int >> dp(m + 1, vector<int >(n + 1));
131+ for (int i = 0; i <= m; ++i) {
132+ dp[ i] [ 0 ] = i;
133+ }
134+ for (int j = 0; j <= n; ++j) {
135+ dp[ 0] [ j ] = j;
136+ }
137+ for (int i = 1; i <= m; ++i) {
138+ for (int j = 1; j <= n; ++j) {
139+ if (word1[ i - 1] == word2[ j - 1] ) {
140+ dp[ i] [ j ] = dp[ i - 1] [ j - 1 ] ;
141+ } else {
142+ dp[ i] [ j ] = min(min(dp[ i - 1] [ j ] , dp[ i] [ j - 1 ] ), dp[ i - 1] [ j - 1 ] ) + 1;
143+ }
144+ }
145+ }
146+ return dp[ m] [ n ] ;
147+ }
148+ };
149+ ```
74150
151+ ### **Go**
152+
153+ ```go
154+ func minDistance(word1 string, word2 string) int {
155+ m, n := len(word1), len(word2)
156+ dp := make([][]int, m+1)
157+ for i := 0; i <= m; i++ {
158+ dp[i] = make([]int, n+1)
159+ dp[i][0] = i
160+ }
161+ for j := 0; j <= n; j++ {
162+ dp[0][j] = j
163+ }
164+ for i := 1; i <= m; i++ {
165+ for j := 1; j <= n; j++ {
166+ if word1[i-1] == word2[j-1] {
167+ dp[i][j] = dp[i-1][j-1]
168+ } else {
169+ dp[i][j] = min(min(dp[i-1][j], dp[i][j-1]), dp[i-1][j-1]) + 1
170+ }
171+ }
172+ }
173+ return dp[m][n]
174+ }
175+
176+ func min(a, b int) int {
177+ if a < b {
178+ return a
179+ }
180+ return b
181+ }
75182```
76183
77184### ** ...**
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