Question
Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 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')
Difficulty:Hard
Category:String, Dynamic-Programming
Analyze
在这道题目中，使用动态规划的方式来求解。设状态f[i][j]表示A[0,i]和B[0,j]之间的最小编辑距离，假设两个string分别为word1和word2.
- 如果
word1[i-1] == word2[j-1], 则f[i][j] = f[i-1][j-1], 这就是表示A[0, i]和B[0,j]之间的编辑距离为f[i][j]. - 如果
word1[i-1] != word2[j-1], 则- 如果
word1[i-1]替换为word2[j-1]，那么f[i][j] = f[i-1][j-1] + 1 - 如果
word1[i-1]后添加word2[j-1]，那么f[i][j] = f[i][j-1] + 1 - 如果
word1[i-1]删除，那么f[i][j] = f[i-1][j] + 1
- 如果
Solution
class Solution {
public:
int minDistance(string word1, string word2) {
const size_t m = word1.length();
const size_t n = word2.length();
int f[m + 1][n + 1];
for (size_t i = 0; i <= m; ++i) {
f[i][0] = i;
}
for (size_t j = 0; j <= n; ++j) {
f[0][j] = j;
}
for (size_t i = 1; i <= m; ++i) {
for (size_t j = 1; j <= n; ++j) {
if (word1[i - 1] == word2[j - 1]) {
f[i][j] = f[i - 1][j - 1];
} else {
f[i][j] = min(f[i - 1][j - 1], min(f[i - 1][j], f[i][j - 1])) + 1;
}
}
}
return f[m][n];
}
};