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:
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.class Solution {
public:
int minDistance(string word1, string word2) {
int m = word1.size(), n = word2.size(), pre;
vector<int> cur(n + 1, 0);
for (int j = 1; j <= n; j++) {
cur[j] = j;
}
for (int i = 1; i <= m; i++) {
pre = cur[0];
cur[0] = i;
for (int j = 1; j <= n; j++) {
int temp = cur[j];
if (word1[i - 1] == word2[j - 1]) {
cur[j] = pre;
} else {
cur[j] = min(pre, min(cur[j - 1], cur[j])) + 1;
}
pre = temp;
}
}
return cur[n];
}
};