题目
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')
链接: https://leetcode-cn.com/problems/edit-distance
- dp[i][j] = dp[i - 1][j - 1] ,Word[i-1]=word2[j-1]
- dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1] , dp[i - 1][j]), dp[i][j - 1]) + 1,Word[i-1]!=word2[j-1]
- for (int j = 0; j <= word2.length(); ++j) {
- dp[0][j] = j;
- }
- for (int i = 0; i <= word1.length(); ++i) {
- dp[i][0] = i;
- }
- class Solution {
- public int minDistance(String word1, String word2) {
- int[][] dp = new int[word1.length() + 1][word2.length() + 1];
- for (int j = 0; j <= word2.length(); ++j) {
- dp[0][j] = j;
- }
- for (int i = 0; i <= word1.length(); ++i) {
- dp[i][0] = i;
- }
- for (int i = 1; i <= word1.length(); ++i) {
- for (int j = 1; j <= word2.length(); ++j) {
- if (word2.charAt(j - 1) == word1.charAt(i - 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[word1.length()][word2.length()];
- }
- }
来源: http://www.bubuko.com/infodetail-3408805.html