{"id":5504,"date":"2019-08-29T23:24:35","date_gmt":"2019-08-30T06:24:35","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=5504"},"modified":"2019-08-29T23:32:55","modified_gmt":"2019-08-30T06:32:55","slug":"leetcode-1143-longest-common-subsequence","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/dynamic-programming\/leetcode-1143-longest-common-subsequence\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1143. Longest Common Subsequence"},"content":{"rendered":"\n

Given two strings text1<\/code> and text2<\/code>, return the length of their longest common subsequence.<\/p>\n\n\n\n

A subsequence<\/em> of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence<\/em> of two strings is a subsequence that is common to both strings.<\/p>\n\n\n\n

If there is no common subsequence, return 0.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

Input:<\/strong> text1 = \"abcde\", text2 = \"ace\" \nOutput:<\/strong> 3  \nExplanation:<\/strong> The longest common subsequence is \"ace\" and its length is 3.\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
Input:<\/strong> text1 = \"abc\", text2 = \"abc\"\nOutput:<\/strong> 3\nExplanation:<\/strong> The longest common subsequence is \"abc\" and its length is 3.\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
Input:<\/strong> text1 = \"abc\", text2 = \"def\"\nOutput:<\/strong> 0\nExplanation:<\/strong> There is no such common subsequence, so the result is 0.\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • 1 <= text1.length <= 1000<\/code><\/li>
  • 1 <= text2.length <= 1000<\/code><\/li>
  • The input strings consist of lowercase English characters only.<\/li><\/ul>\n\n\n\n
    Solution: DP<\/strong><\/p>\n\n\n\n
    Use dp[i][j] to represent the length of longest common sub-sequence of text1[0:i] and text2[0:j]
    dp[i][j] = dp[i – 1][j – 1]  + 1 if text1[i – 1] == text2[j – 1] else max(dp[i][j – 1], dp[i – 1][j])<\/p>\n\n\n\n
    Time complexity: O(mn)
    Space complexity: O(mn) -> O(n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua, 16ms 14.9 MB\nclass Solution {\npublic:\n  int longestCommonSubsequence(string text1, string text2) {\n    int m = text1.length();\n    int n = text2.length();\n    vector> dp(m + 1, vector(n + 1));\n    for (int i = 0; i < m; ++i)\n      for (int j = 0; j < n; ++j)\n        if (text1[i] == text2[j])\n          dp[i + 1][j + 1] = dp[i][j] + 1;\n        else\n          dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j]);      \n    return dp[m][n];\n  }\n};\n<\/pre>\n\n<\/div>
    C++\/V2<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua, 8ms, 8.8MB\nclass Solution {\npublic:\n  int longestCommonSubsequence(string text1, string text2) {\n    int m = text1.length();\n    int n = text2.length();    \n    vector dp(n + 1);    \n    for (int i = 0; i < m; ++i) {\n      int prev = 0; \/\/ dp[i][j]\n      for (int j = 0; j < n; ++j) {        \n        int curr = dp[j + 1]; \/\/ dp[i][j + 1]\n        if (text1[i] == text2[j])\n          \/\/ dp[i + 1][j + 1] = dp[i][j] + 1\n          dp[j + 1] = prev + 1; \n        else\n          \/\/ dp[i + 1][j + 1] = max(dp[i][j + 1], dp[i + 1][j])\n          dp[j + 1] = max(curr, dp[j]);\n        prev = curr;\n      }    \n    }    \n    return dp[n]; \/\/ dp[m][n]\n  }\n};\n<\/pre>\n\n<\/div>
    C++\/V3<\/h2>\n
    \n\n
    \n\/\/ Author: Huahua\nclass Solution {\npublic:\n  int longestCommonSubsequence(string text1, string text2) {\n    int m = text1.length();\n    int n = text2.length();    \n    vector dp1(n + 1);\n    vector dp2(n + 1);\n    for (int i = 0; i < m; ++i) {      \n      for (int j = 0; j < n; ++j)\n        if (text1[i] == text2[j])\n          dp2[j + 1] = dp1[j] + 1; \n        else          \n          dp2[j + 1] = max(dp1[j + 1], dp2[j]);              \n      swap(dp1, dp2);\n    }    \n    return dp1[n];\n  }\n};\n<\/pre>\n<\/div><\/div>\n\n\n\n
    <\/p>\n","protected":false},"excerpt":{"rendered":"
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