题目：
Given two strings?word1?and?word2, return?the minimum number of operations required to convert?word1?to?word2.
You have the following three operations permitted on a word:
- Insert a character
- Delete a character
- Replace a character
Example 1:
Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution" Output: 5 Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')
Constraints:
0 <= word1.length, word2.length <= 500word1?and?word2?consist of lowercase English letters.
思路：
dp题，相当于583. Delete Operation for Two Strings的follow up题，只是变换方式不同。同样，dp[i][j]即使字符串s1 [0, i - 1]截取的字符串和s2 [0, j - 1]截取的字符串相等的情况，需要进行多少次操作。如果当前两个字母相等的话，那就不用操作了，直接dp[i][j] = dp[i - 1][j - 1]即可。如果不同，我们以s1为基点考虑，因为s1和s2是相对的，在s1插入相当于在s2删除，因此我们以s1为基点考虑，就不用再考虑一次在s2上的操作了。如果不同，dp是三种情况，在s1删除一个字符，即s1[i - 1]删掉，那么就来自s1[i - 2]的结果，即dp[i - 1][j]；插入一个字符，相当于在s2删除一个字符，即s2[j - 1]删掉，那么就来自dp[i][j - 1] + 1；替换一个字符，那就相当于不删除，则来自dp[i - 1][j - 1] + 1。初始化和583是一样的，从dp定义出发，对于f[i][0]来说，即s1的子串[0, i]删除几次才能和空字符串相同，显然是i次，对于f[0][j]同理。
代码：
class Solution {
public:
? ? int minDistance(string s1, string s2) {
? ? ? ? int m = s1.size(), n = s2.size();
? ? ? ? vector<vector<int>> f(m + 1, vector<int>(n + 1));
? ? ? ? for (int i = 1; i < m + 1; i++)
? ? ? ? ? ? f[i][0] = i;
? ? ? ? for (int j = 1; j < n + 1; j++)
? ? ? ? ? ? f[0][j] = j;
? ? ? ? for (int i = 1; i < m + 1; i++) {
? ? ? ? ? ? for (int j = 1; j < n + 1; j++) {
? ? ? ? ? ? ? ? if (s1[i - 1] == s2[j - 1])
? ? ? ? ? ? ? ? ? ? f[i][j] = f[i - 1][j - 1];
? ? ? ? ? ? ? ? else
? ? ? ? ? ? ? ? ? ? f[i][j] = min({ f[i - 1][j], f[i][j - 1], f[i - 1][j - 1] }) + 1;
? ? ? ? ? ? }
? ? ? ? }
? ? ? ? return f[m][n];
? ? }
};
