leetcode 72. 编辑距离

给你两个单词 word1 和 word2,请你计算出将 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')
 

提示:

0 <= word1.length, word2.length <= 500
word1 和 word2 由小写英文字母组成

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/edit-distance

采用动态规划

class Solution {
    public int minDistance(String word1, String word2) {
         int l1 = word1.length();
        int l2 = word2.length();
        if ("".equals(word1)) {
            return l2;
        }
        if ("".equals(word2)) {
            return l1;
        }
        char[] cs1 = word1.toCharArray();
        char[] cs2 = word2.toCharArray();
        int[] dp = new int[l2 + 1];
        for (int i = 0; i <= l2; i++) {
            dp[i] = i;
        }
        int sum;
        int item;
        for (int i = 1; i <= l1; i++) {
            item = dp[0];
            dp[0] = i;
            for (int j = 1; j <= l2; j++) {
                sum = dp[j];
                if (cs1[i - 1] == cs2[j - 1]) {
                    dp[j] = item;
                } else {
                    dp[j] = Math.min(Math.min(item, dp[j - 1]), dp[j]) + 1;
                }
                item = sum;
            }
        }
        return dp[l2];
    }
}

posted @ 2021-04-14 14:57  旺仔古李  阅读(40)  评论(0编辑  收藏  举报