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EditDistance.java
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58 lines (57 loc) · 1.78 KB
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package com.yangchd.leetcode.hard;
/**
* @author yangchd 2018/10/31
*
* 72.Edit Distance
* 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')
*/
public class EditDistance {
/**
* 经典的DP算法
* 用二阶矩阵来记录每一步的可能
*/
class Solution {
public int minDistance(String word1, String word2) {
int[][] dp = new int[word1.length() + 1][word2.length() + 1];
for (int i = 0; i < word1.length() + 1; i++) {
dp[i][0] = i;
}
for (int j = 0; j < word2.length() + 1; j++) {
dp[0][j] = j;
}
for (int m = 1; m < word1.length() + 1; m++) {
for (int n = 1; n < word2.length() + 1; n++) {
if (word1.charAt(m - 1) == word2.charAt(n - 1)) {
dp[m][n] = dp[m - 1][n - 1];
} else {
dp[m][n] = Math.min(dp[m - 1][n - 1], Math.min(dp[m][n - 1], dp[m - 1][n])) + 1;
}
}
}
return dp[word1.length()][word2.length()];
}
}
}