From 852b6a85af7c3d1a96af1a26b7056c6863653751 Mon Sep 17 00:00:00 2001 From: huangge1199 Date: Wed, 11 Aug 2021 11:22:44 +0800 Subject: [PATCH] =?UTF-8?q?72:=E7=BC=96=E8=BE=91=E8=B7=9D=E7=A6=BB?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../java/leetcode/editor/cn/EditDistance.java | 80 +++++++++++++++++++ .../java/leetcode/editor/cn/EditDistance.md | 45 +++++++++++ 2 files changed, 125 insertions(+) create mode 100644 src/main/java/leetcode/editor/cn/EditDistance.java create mode 100644 src/main/java/leetcode/editor/cn/EditDistance.md diff --git a/src/main/java/leetcode/editor/cn/EditDistance.java b/src/main/java/leetcode/editor/cn/EditDistance.java new file mode 100644 index 0000000..2178407 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/EditDistance.java @@ -0,0 +1,80 @@ +//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 +// +// Related Topics 字符串 动态规划 +// 👍 1750 👎 0 + +package leetcode.editor.cn; + +//72:编辑距离 +class EditDistance { + public static void main(String[] args) { + //测试代码 + Solution solution = new EditDistance().new Solution(); + System.out.println(solution.minDistance("horse", "ros")); + System.out.println(solution.minDistance("intention", "execution")); + } + + //力扣代码 + //leetcode submit region begin(Prohibit modification and deletion) + class Solution { + public int minDistance(String word1, String word2) { + int[][] dp = new int[word1.length() + 1][word2.length() + 1]; + for (int i = 0; i < word1.length() + 1; i++) { + dp[i][0] = i; + } + for (int i = 0; i < word2.length() + 1; i++) { + dp[0][i] = i; + } + for (int i = 1; i < word1.length() + 1; i++) { + for (int j = 1; j < word2.length() + 1; j++) { + 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]); + } + } + return dp[word1.length()][word2.length()]; + } + } +//leetcode submit region end(Prohibit modification and deletion) + +} \ No newline at end of file diff --git a/src/main/java/leetcode/editor/cn/EditDistance.md b/src/main/java/leetcode/editor/cn/EditDistance.md new file mode 100644 index 0000000..eeb8840 --- /dev/null +++ b/src/main/java/leetcode/editor/cn/EditDistance.md @@ -0,0 +1,45 @@ +

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:

+ + + +

 

+ +

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:

+ + +
Related Topics
  • 字符串
  • 动态规划
  • \n
  • 👍 1750
  • 👎 0
  • \ No newline at end of file