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  • letcode1143 Longest Common Subsequence

    letcode1143 Longest Common Subsequence

    Longest Common Subsequence
    Given two strings text1 and text2, return the length of their longest common subsequence.

    A subsequence 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 of two strings is a subsequence that is common to both strings.

    If there is no common subsequence, return 0.

    Example 1:

    Input: text1 = "abcde", text2 = "ace"
    Output: 3
    Explanation: The longest common subsequence is "ace" and its length is 3.
    Example 2:

    Input: text1 = "abc", text2 = "abc"
    Output: 3
    Explanation: The longest common subsequence is "abc" and its length is 3.
    Example 3:

    Input: text1 = "abc", text2 = "def"
    Output: 0
    Explanation: There is no such common subsequence, so the result is 0.

    Constraints:

    1 <= text1.length <= 1000
    1 <= text2.length <= 1000
    The input strings consist of lowercase English characters only.
    Hide Hint #1
    Try dynamic programming. DP[i][j] represents the longest common subsequence of text1[0 ... i] & text2[0 ... j].
    Hide Hint #2
    DP[i][j] = DP[i - 1][j - 1] + 1 , if text1[i] == text2[j] DP[i][j] = max(DP[i - 1][j], DP[i][j - 1]) , otherwise

    solution

    approach1 dp

    class Solution {
    public:
        int longestCommonSubsequence(string text1, string text2) {
            int a=text1.length();
            int b=text2.length();
            int dp[a+1][b+1];
            for(int i =0;i<a;i++)
                for(int j=0;j<b;j++)
                {
                    if(i==0)
                    {
                        if(text2.find(text1[0])<=j)
                            dp[i][j]=1;
                        else
                            dp[i][j]=0;
                        continue;
                    }
                    if(j==0)
                    {
                        if(text1.find(text2[0])<=i)
                            dp[i][j]=1;
                        else
                            dp[i][j]=0;
                        continue;
                    }
                    if(text1[i] == text2[j])
                        dp[i][j] = dp[i - 1][j - 1] + 1;
                    else
                        dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]);
                }
            return dp[a-1][b-1];
        }
    };
    

    参考链接

    lletcode

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  • 原文地址:https://www.cnblogs.com/qwfand/p/12780337.html
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