动态规划之编辑距离

编辑距离这道题本质上也是遍历所有问题分支的题目，他有三个选择，插入删除替换

dp[i][j] 这个定义代表 前i个字匹配前j个字符对应的编辑距离

dp[i][j-1] +1 代表 ，在word2后增加一个字符，word1前i个字符转换到word2前j个字符的距离为dp[i][j-1]+1 。这是word2增，

dp[i-1][j]+1 代表 word2 删除

dp[i-1][j-1]+1 替换

/**
 * <p>给你两个单词 <code>word1</code> 和 <code>word2</code>， <em>请返回将 <code>word1</code> 转换成 <code>word2</code> 所使用的最少操作数</em>  。</p>
 *
 * <p>你可以对一个单词进行如下三种操作：</p>
 *
 * <ul>
 * <li>插入一个字符</li>
 * <li>删除一个字符</li>
 * <li>替换一个字符</li>
 * </ul>
 *
 * <p> </p>
 *
 * <p><strong>示例 1：</strong></p>
 *
 * <pre>
 * <strong>输入：</strong>word1 = "horse", word2 = "ros"
 * <strong>输出：</strong>3
 * <strong>解释：</strong>
 * horse -> rorse (将 'h' 替换为 'r')
 * rorse -> rose (删除 'r')
 * rose -> ros (删除 'e')
 * </pre>
 *
 * <p><strong>示例 2：</strong></p>
 *
 * <pre>
 * <strong>输入：</strong>word1 = "intention", word2 = "execution"
 * <strong>输出：</strong>5
 * <strong>解释：</strong>
 * 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')
 * </pre>
 *
 * <p> </p>
 *
 * <p><strong>提示：</strong></p>
 *
 * <ul>
 * <li><code>0 <= word1.length, word2.length <= 500</code></li>
 * <li><code>word1</code> 和 <code>word2</code> 由小写英文字母组成</li>
 * </ul>
 * <div><div>Related Topics</div><div><li>字符串</li><li>动态规划</li></div></div><br><div><li>👍 2458</li><li>👎 0</li></div>
 */

//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length();
        int n = word2.length();

        int[][] dp = new int[m+1][n+1];
        for (int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j;
        }
        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] + 1, Math.min(dp[i][j - 1] + 1, dp[i - 1][j - 1] + 1));
                }
            }
        }
        return dp[m][n];
    }
}
//leetcode submit region end(Prohibit modification and deletion)