80 lines
2.2 KiB
Java
80 lines
2.2 KiB
Java
//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 。
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//
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// 你可以对一个单词进行如下三种操作：
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//
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//
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// 插入一个字符
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// 删除一个字符
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// 替换一个字符
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//
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//
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//
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//
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// 示例 1：
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//
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//
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//输入：word1 = "horse", word2 = "ros"
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//输出：3
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//解释：
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//horse -> rorse (将 'h' 替换为 'r')
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//rorse -> rose (删除 'r')
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//rose -> ros (删除 'e')
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//
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//
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// 示例 2：
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//
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//
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//输入：word1 = "intention", word2 = "execution"
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//输出：5
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//解释：
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//intention -> inention (删除 't')
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//inention -> enention (将 'i' 替换为 'e')
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//enention -> exention (将 'n' 替换为 'x')
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//exention -> exection (将 'n' 替换为 'c')
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//exection -> execution (插入 'u')
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//
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//
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//
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//
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// 提示：
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//
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//
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// 0 <= word1.length, word2.length <= 500
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// word1 和 word2 由小写英文字母组成
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//
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// Related Topics 字符串 动态规划
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// 👍 1750 👎 0
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package leetcode.editor.cn;
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//72:编辑距离
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class EditDistance {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new EditDistance().new Solution();
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System.out.println(solution.minDistance("horse", "ros"));
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System.out.println(solution.minDistance("intention", "execution"));
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public int minDistance(String word1, String word2) {
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int[][] dp = new int[word1.length() + 1][word2.length() + 1];
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for (int i = 0; i < word1.length() + 1; i++) {
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dp[i][0] = i;
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}
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for (int i = 0; i < word2.length() + 1; i++) {
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dp[0][i] = i;
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}
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for (int i = 1; i < word1.length() + 1; i++) {
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for (int j = 1; j < word2.length() + 1; j++) {
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dp[i][j] = 1 + Math.min(Math.min(dp[i - 1][j], dp[i][j - 1]), word2.charAt(j - 1) == word1.charAt(i - 1) ? dp[i - 1][j - 1] - 1 : dp[i - 1][j - 1]);
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}
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}
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return dp[word1.length()][word2.length()];
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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} |