Java解决LeetCode72题 Edit Distance

题目描述

地址 : https://leetcode.com/problems/edit-distance/description/

题目描述

思路

  • 使用dp[i][j]用来表示word10~i-1word20~j-1的最小编辑距离
  • 我们可以知道边界情况:dp[i][0] = idp[0][j] = j,代表从 "" 变为 dp[0~i-1]dp[0][0~j-1] 所需要的次数

同时对于两个字符串的子串,都能分为最后一个字符相等或者不等的情况:

  • 如果word1[i-1] == word2[j-1]dp[i][j] = dp[i-1][j-1]
  • 如果word1[i-1] != word2[j-1]
    • 向word1插入:dp[i][j] = dp[i][j-1] + 1
    • 从word1删除:dp[i][j] = dp[i-1][j] + 1
    • 替换word1元素:dp[i][j] = dp[i-1][j-1] + 1
 public int minDistance(String word1, String word2) {
    int n = word1.length();
    int m = word2.length();
    int[][] dp = new int[n + 1][m + 1];
    for (int i = 0; i < m + 1; i++) {
        dp[0][i] = i;
    }
    for (int i = 0; i < n + 1; i++) {
        dp[i][0] = i;
    }
    for (int i = 1; i < n + 1; i++) {
        for (int j = 1; j < m + 1; 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], dp[i][j - 1]), dp[i - 1][j - 1]) + 1;
            }
        }
    }
    return dp[n][m];
}
posted @ 2019-02-08 17:29 morethink 阅读(...) 评论(...) 编辑 收藏