LeetCode-in-Cpp
72. Edit Distance
Hard
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 <= 500word1andword2consist of lowercase English letters.
Solution
#include <vector>
#include <string>
#include <algorithm>
class Solution {
public:
int minDistance(const std::string& w1, const std::string& w2) {
int n1 = w1.length();
int n2 = w2.length();
if (n2 > n1) {
return minDistance(w2, w1);
}
std::vector<int> dp(n2 + 1);
for (int j = 0; j <= n2; j++) {
dp[j] = j;
}
for (int i = 1; i <= n1; i++) {
int pre = dp[0];
dp[0] = i;
for (int j = 1; j <= n2; j++) {
int tmp = dp[j];
dp[j] = (w1[i - 1] != w2[j - 1])
? 1 + std::min(pre, std::min(dp[j], dp[j - 1]))
: pre;
pre = tmp;
}
}
return dp[n2];
}
};