/*
 * @lc app=leetcode.cn id=72 lang=java
 *
 * [72] 编辑距离
 *
 * https://leetcode-cn.com/problems/edit-distance/description/
 *
 * algorithms
 * Hard (59.31%)
 * Likes:    912
 * Dislikes: 0
 * Total Accepted:    62K
 * Total Submissions: 104.5K
 * Testcase Example:  '"horse"\n"ros"'
 *
 * 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')
 * 
 * 
 */

// @lc code=start
class Solution {
    public int minDistance(String word1, String word2) {
        //状态转移方程  
        //word1(i-1)==word2(j-1) >> dp[i][j]=dp[i-1][j-1]
        //else >> dp[i][j]=min(dp[i][j-1],d[i-1][j],d[i-1][j-1])+1

        int m = word1.length();
        int n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int i = 1; i <= n; i++) {
            dp[0][i] = i;
        }

        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = Math.min(dp[i - 1][j], Math.min(dp[i][j - 1], dp[i - 1][j - 1])) + 1;
                }
            }
        }
        return dp[m][n];
    }
}
// @lc code=end