• LeetCode 1143. Longest Common Subsequence


    原题链接在这里:https://leetcode.com/problems/longest-common-subsequence/

    题目:

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

    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.

    题解:

    dp[i][j] stands for length of LCS between text1 up to i and text2 up to j.

    If text1.charAt(i) == text2.charAt(j), dp[i][j] = dp[i-1][j-1] + 1.

    Otherwise, dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j]).

    如果不放心的话,就直接取上述三个的最小.

    Time Complexity: O(m*n). m = text1.length. n = text2.length.

    Space: O(m*n).

    AC Java: 

     1 class Solution {
     2     public int longestCommonSubsequence(String text1, String text2) {
     3         if(text1 == null || text1.length() == 0 || text2 == null || text2.length() == 0){
     4             return 0;
     5         }
     6         
     7         int m = text1.length();
     8         int n = text2.length();
     9         int [][] dp = new int[m+1][n+1];
    10         for(int i = 0; i<m; i++){
    11             for(int j = 0; j<n; j++){
    12                 dp[i+1][j+1] = Math.max(dp[i][j+1], dp[i+1][j]);
    13                 if(text1.charAt(i) == text2.charAt(j)){
    14                     dp[i+1][j+1] = Math.max(dp[i+1][j+1], dp[i][j]+1);
    15                 }                
    16             }
    17         }
    18         
    19         return dp[m][n];
    20     }
    21 }

    类似Longest PalindromicMaximum Length of Repeated Subarray.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/11450819.html
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