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LeetCode-72-Edit-Distance.java
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LeetCode-72-Edit-Distance.java
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/*
https://discuss.leetcode.com/topic/17639/20ms-detailed-explained-c-solutions-o-n-space
https://www.cnblogs.com/grandyang/p/4344107.html
https://www.youtube.com/watch?v=Q4i_rqON2-E
Example:
a b c d e f
a z c e d
b -> z
delete d
f -> d
Total 3 edits to convert "abcdef" to "azced"
Three Operations:
Case 1: word1[i] = word2[j]
f(i, j) = f(i - 1, j - 1)
Case 2: word1[i] != word2[j], then we must either insert, delete or replace, whichever is cheaper
1. f(i, j) = f(i, j - 1) + 1 - insert in word2
2. f(i, j) = f(i - 1, j) + 1 - delete in word1
3. f(i, j) = f(i - 1, j - 1) + 1 - replace the char
Matrix
a b c d e f
0 1 1 3 4 5 6
a 1 0 1 2 3 4 5
z 2 1 1 2 3 4 5
c 3 2 2 1 2 3 4
e 4 3 3 2 2 2 4
d 5 4 4 3 3 3 3
*/
public class Solution {
// public int minDistance(String word1, String word2) {
// int m = word1.length(), n = word2.length();
// // state matrix
// int[][] dp = new int[m + 1][n + 1];
// // init matrix
// for(int i = 1; i <= m; i++) dp[i][0] = i;
// for(int j = 1; j <= n; j++) dp[0][j] = j;
// // calculate matrix
// for(int i = 0; i < m; i++){
// for(int j = 0; j < n; j++){
// if(word1.charAt(i) == word2.charAt(j)) dp[i + 1][j + 1] = dp[i][j];
// else dp[i + 1][j + 1] = Math.min(dp[i][j] + 1, Math.min(dp[i][j + 1] + 1, dp[i + 1][j] + 1));
// }
// }
// return dp[m][n];
// }
// similar to the above one
public int minDistance(String word1, String word2) {
int m = word1.length(), n = word2.length();
// subproblem
int[][] dp = new int[m + 1][n + 1];
// init
for (int i = 1; i <= m; i++) dp[i][0] = i;
for (int j = 1; j <= n; j++) dp[0][j] = j;
// recurrence relation
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i - 1][j], dp[i][j - 1])) + 1;
}
}
}
// ans
return dp[m][n];
}
}