72:编辑距离

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huangge1199 2021-08-11 11:22:44 +08:00
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//给你两个单词 word1 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 ( '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 ( 'i' 替换为 'e')
//enention -> exention ( 'n' 替换为 'x')
//exention -> exection ( 'n' 替换为 'c')
//exection -> execution (插入 'u')
//
//
//
//
// 提示
//
//
// 0 <= word1.length, word2.length <= 500
// word1 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)
}

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<p>给你两个单词 <code>word1</code> 和 <code>word2</code>,请你计算出将 <code>word1</code> 转换成 <code>word2</code><em> </em>所使用的最少操作数 。</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><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>\n<div><li>👍 1750</li><li>👎 0</li></div>