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EditDistance.java
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EditDistance.java
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package com.hncboy;
/**
* @author hncboy
* @date 2019/10/17 16:42
* @description 72.编辑距离
*
* 给定两个单词 word1 和 word2,计算出将 word1 转换成 word2 所使用的最少操作数 。
* 你可以对一个单词进行如下三种操作:
*
* 插入一个字符
* 删除一个字符
* 替换一个字符
*
* 示例 1:
* 输入: word1 = "horse", word2 = "ros"
* 输出: 3
* 解释:
* horse -> rorse (将 'h' 替换为 'r')
* rorse -> rose (删除 'r')
* rose -> ros (删除 'e')
*
* 示例 2:
* 输入: word1 = "intention", word2 = "execution"
* 输出: 5
* 解释:
* intention -> inention (删除 't')
* inention -> enention (将 'i' 替换为 'e')
* enention -> exention (将 'n' 替换为 'x')
* exention -> exection (将 'n' 替换为 'c')
* exection -> execution (插入 'u')
*/
public class EditDistance {
public static void main(String[] args) {
EditDistance e = new EditDistance();
System.out.println(e.minDistance("horse", "ros"));
System.out.println(e.minDistance("intention", "execution"));
System.out.println(e.minDistance("", "a"));
System.out.println(e.minDistance("a", ""));
System.out.println(e.minDistance("a", "a"));
System.out.println(e.minDistance("dinitrophenylhydrazine", "dimethylhydrazine"));
}
private int minDistance(String word1, String word2) {
int length1 = word1.length();
int length2 = word2.length();
int[][] dp = new int[length1 + 1][length2 + 1];
for (int i = 0; i <= length1; i++) {
dp[i][0] = i;
}
for (int j = 0; j <= length2; j++) {
dp[0][j] = j;
}
for (int i = 1; i <= length1; i++) {
for (int j = 1; j <= length2; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i - 1][j]), dp[i][j - 1]) + 1;
}
}
}
return dp[length1][length2];
}
}