mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-09-04 15:01:40 +08:00
update
This commit is contained in:
@@ -12,7 +12,7 @@
|
||||
"translatedContent": "<p>给定两个字符串 <code>text1</code> 和 <code>text2</code>,返回这两个字符串的最长 <strong>公共子序列</strong> 的长度。如果不存在 <strong>公共子序列</strong> ,返回 <code>0</code> 。</p>\n\n<p>一个字符串的 <strong>子序列</strong><em> </em>是指这样一个新的字符串:它是由原字符串在不改变字符的相对顺序的情况下删除某些字符(也可以不删除任何字符)后组成的新字符串。</p>\n\n<ul>\n\t<li>例如,<code>\"ace\"</code> 是 <code>\"abcde\"</code> 的子序列,但 <code>\"aec\"</code> 不是 <code>\"abcde\"</code> 的子序列。</li>\n</ul>\n\n<p>两个字符串的 <strong>公共子序列</strong> 是这两个字符串所共同拥有的子序列。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>text1 = \"abcde\", text2 = \"ace\" \n<strong>输出:</strong>3 \n<strong>解释:</strong>最长公共子序列是 \"ace\" ,它的长度为 3 。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>text1 = \"abc\", text2 = \"abc\"\n<strong>输出:</strong>3\n<strong>解释:</strong>最长公共子序列是 \"abc\" ,它的长度为 3 。\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>text1 = \"abc\", text2 = \"def\"\n<strong>输出:</strong>0\n<strong>解释:</strong>两个字符串没有公共子序列,返回 0 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= text1.length, text2.length <= 1000</code></li>\n\t<li><code>text1</code> 和 <code>text2</code> 仅由小写英文字符组成。</li>\n</ul>\n",
|
||||
"isPaidOnly": false,
|
||||
"difficulty": "Medium",
|
||||
"likes": 905,
|
||||
"likes": 909,
|
||||
"dislikes": 0,
|
||||
"isLiked": null,
|
||||
"similarQuestions": "[]",
|
||||
@@ -143,7 +143,7 @@
|
||||
"__typename": "CodeSnippetNode"
|
||||
}
|
||||
],
|
||||
"stats": "{\"totalAccepted\": \"213.4K\", \"totalSubmission\": \"332.7K\", \"totalAcceptedRaw\": 213403, \"totalSubmissionRaw\": 332664, \"acRate\": \"64.1%\"}",
|
||||
"stats": "{\"totalAccepted\": \"214.4K\", \"totalSubmission\": \"334.1K\", \"totalAcceptedRaw\": 214370, \"totalSubmissionRaw\": 334147, \"acRate\": \"64.2%\"}",
|
||||
"hints": [
|
||||
"Try dynamic programming. \r\nDP[i][j] represents the longest common subsequence of text1[0 ... i] & text2[0 ... j].",
|
||||
"DP[i][j] = DP[i - 1][j - 1] + 1 , if text1[i] == text2[j]\r\nDP[i][j] = max(DP[i - 1][j], DP[i][j - 1]) , otherwise"
|
||||
|
Reference in New Issue
Block a user